Question 13 Marks
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/ have the same energy lists:
- $\text{n}=4,\text{l}=2,\text{m}_\text{l}=-2,\text{m}_{\text{s}}=-\frac{1}{2}$
- $\text{n}=3,\text{l}=2,\text{m}_\text{l}=1,\text{m}_{\text{s}}=+\frac{1}{2}$
- $\text{n}=4,\text{l}=1,\text{m}_\text{l}=0,\text{m}_{\text{s}}=+\frac{1}{2}$
- $\text{n}=3,\text{l}=2,\text{m}_\text{l}=-2,\text{m}_{\text{s}}=-\frac{1}{2}$
- $\text{n}=3,\text{l}=1,\text{m}_\text{l}=-1,\text{m}_{\text{s}}=+\frac{1}{2}$
- $\text{n}=4,\text{l}=1,\text{m}_\text{l}=0,\text{m}_{\text{s}}=+\frac{1}{2}$
AnswerFor n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.
Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1(4d).
View full question & answer→Question 23 Marks
Calculate the wavelength of an electron moving with a velocity of $2.05 \times 107 \mathrm{~ms}^{-1}$.
AnswerAccording to de Broglie’s equation,
$\lambda=\frac{\text{h}}{\text{mv}}$
Where,
$\lambda$ = wavelength of moving particle
m = mass of particle
v = velocity of particle
h = Planck’s constant
Substituting the values in the expression of $\lambda:$
$\lambda=\frac{6.626\times10^{-34}\text{Js}}{(9.10939\times10^{-31}\text{kg})(2.05\times10^7\text{ms}^{-1})}$
$\lambda=3.548\times10^{-11}\text{m}$
Hence, the wavelength of the electron moving with a velocity of $2.05 \times 107 \mathrm{~ms}^{-1}$ is $3.548 \times 10^{-11} \mathrm{~m}$.
View full question & answer→Question 33 Marks
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.
AnswerLet the number of electrons in the ion = x
Then, number of neutrons $=\text{x}+\Big(\frac{30.4\text{x}}{100}\Big)=1.304\text{x}$
Number of electrons in the neutral atom = x + 3 (ion possesses 3 units of positive charge)
$\therefore$ Number of protons = x + 3
Mass number = No. of protons + No. of neutrons
$\therefore$ 1.304 x + x + 3 = 56
⇒ 2.304 x = 53
⇒ x = 23
$\therefore$ No. of protons = Atomic no. = x + 3 = 23 + 3 = 26
The symbol of the ion is $\text{ }^{56}_{29}\text{Fe}^{+3}$
View full question & answer→Question 43 Marks
The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?
Answer Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom.
The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13). Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.
View full question & answer→Question 53 Marks
Find
- The total number and.
- The total mass of protons in 34mg of $NH_3$ at $STP.$
Will the answer change if the temperature and pressure are changed? Answer1 mol of $NH_3= 17g NH_3= 6.022 \times 10^{23}$ molecules of $NH_3$
1 atom of $NH_3$ contains = 7 + 3 = 10 protons
$\therefore$ The number of protons in 1 mol of $NH_3= 6.022 \times 10^{24}$ protons.
Number of protons in 34mg of $NH_3$ $=\frac{(6.022\times10^{24}\times34)}{17\times1000}=1.2044\times10^{22}\text{ protons.}$
Mass of one proton $= 1.6726 \times 10^{-27}kg$
$\therefore$ Mass of $1.2044 \times 10^{22}$ protons $= (1.6726 \times 10^{-27}) \times (1.2044\times 10^{22})kg = 2.0145 \times 10^{-5}kg.$
No, there will be no effect of temperature and pressure.
View full question & answer→Question 63 Marks
Give the number of electrons in the species $\text{H}^+_2,\text{H}_2$ and $\text{O}^+_2$
Answer$\text{H}^+_2:$
Number of electrons present in hydrogen molecule ($\text{H}_2$) = 1 + 1 = 2
$\therefore$ Number of electrons in $\text{H}^+_2 = 2 – 1 = 1$
$\text{H}_2:$
Number of electrons in $\text{H}_2= 1 + 1 = 2$
$\text{O}^+_2:$
Number of electrons present in oxygen molecule ($\text{O}_2$) = 8 + 8 = 16
$\therefore$ Number of electrons in $\text{O}^+_2= 16 – 1 = 15$
View full question & answer→Question 73 Marks
The mass of an electron is $9.1 \times 10^{-31} \mathrm{~kg}$. If its K.E. is $3.0 \times 10^{-25} \mathrm{~J}$, calculate its wavelength.
AnswerFrom de Broglie’s equation,
$\lambda=\frac{\text{h}}{\text{mv}}$
Given,
Kinetic energy (K.E) of the electron = $3.0 \times 10^{-25} \mathrm{~J}$
Since $\text{K.E}=\frac{1}{2}\text{mv}^2$
$\therefore\text{Velocity (v)}=\sqrt{\frac{2\text{K.E}}{\text{m}}}$
$=\sqrt{\frac{2(3.0\times10^{-25}\text{J})}{9.10939\times10^{-31}\text{kg}}}$
$=\sqrt{6.5866\times10^4}$
$\text{v}=811.579\text{ ms}^{-1}$
Substituting the value in the expression of $\lambda:$
View full question & answer→Question 83 Marks
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is $1.6 \times 10^6 \mathrm{~ms}^{-1}$, calculate de Broglie wavelength associated with this electron.
AnswerFrom de Broglie’s equation,
$\lambda=\frac{\text{h}}{\text{mv}}$
$\lambda=\frac{6.626\times10^{-34}\text{Js}}{(9.10939\times10^{-31}\text{kg})(1.6\times10^6\text{ms}^{-1})}$
$=4.55\times10^{-10}\text{m}$
$\lambda=455\text{pm}$
$\therefore$ De Broglie’s wavelength associated with the electron is 455pm.
View full question & answer→Question 93 Marks
Which atoms are indicated by the following configurations?
- $[He] 2s^1.$
- $[Ne] 3s^2 3p^3.$
- $[Ar] 4s^2 3d^1.$
Answer
- $[He] 2s^1$
The electronic configuration of the element is $[He] 2s^1 = 1s^2 2s^1.$
$\therefore$ Atomic number of the element $= 3$
Hence, the element with the electronic configuration $[He] 2s^1$ is lithium (Li).
- $[Ne] 3s^2 3p^3$
The electronic configuration of the element is $[Ne] 3s^2 3p^3= 1s^2 2s^2 2p^6 3s^2 3p^3.$
$\therefore$ Atomic number of the element $= 15$
Hence, the element with the electronic configuration $[Ne] 3s^2 3p^3$ is phosphorus (P).
- $[Ar] 4s^2 3d^1$
The electronic configuration of the element is $[Ar] 4s^2 3d^1= 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1.$
$\therefore$ Atomic number of the element $= 21$
Hence, the element with the electronic configuration [Ar] $4s^2 3d^1$ is scandium $(Sc).$ View full question & answer→Question 103 Marks
Calculate the mass and charge of one mole of electrons.
Answer$\text { Mass of one electron }=9.10939 \times 10^{-31} \mathrm{~kg}$
$\text { Mass of one mole of electron }=\left(6.022 \times 10^{23}\right) \times\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)$
$=5.48 \times 10^{-7} \mathrm{~kg}$
Charge on one electron $=1.6022 \times 10^{-19}$ coulomb
Charge on one mole of electron $=\left(1.6022 \times 10^{-19} \mathrm{C}\right)\left(6.022 \times 10^{23}\right)$
$=9.65 \times 10^4 \mathrm{C}$
View full question & answer→Question 113 Marks
Find energy of each of the photons which,
Have wavelength of 0.50 $\mathring{\text{A}}.$
AnswerEnergy (E) of a photon having wavelength $(\lambda)$ is given by the expression,
$\text{E}=\frac{\text{hc}}{\lambda}$
h = Planck’s constant = $6.626 \times 10^{–34}Js$
c = velocity of light in vacuum = $3 \times 10^8m/s$
Substituting the values in the given expression of E:
$\text{E}=\frac{(6.626\times10^{-34})(3\times10^8)}{0.50\times10^{-10}}=3.976\times10^{-15}\text{J}$
$\therefore\text{E}=3.98\times10^{-15}\text{J}$
View full question & answer→Question 123 Marks
An ion with mass number 37 possesses one unit of negative charge. If the ion conatins 11.1% more neutrons than the electrons, find the symbol of the ion.
AnswerLet the number of electrons in the ion carrying a negative charge be x.
Then,
Number of neutrons present
= x + 11.1% of x
= x + 0.111 x
= 1.111 x
Number of electrons in the neutral atom = (x – 1)
(When an ion carries a negative charge, it carries an extra electron)
$\therefore$ Number of protons in the neutral atom = x – 1
Given,
Mass number of the ion = 37
$\therefore$ (x – 1) + 1.111 x = 37
2.111 x = 38
x = 18
$\therefore$ The symbol of the ion is $\text{ }^{37}_{17}\text{Cl}^-.$
View full question & answer→Question 133 Marks
Electromagnetic radiation of wavelength 242nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in $\mathrm{~kJ} \mathrm{~mol}^{-1}$.
AnswerEnergy of sodium (E) $=\frac{\text{N}_{\text{A}}\text{hc}}{\lambda}$
$=\frac{(6.023\times10^{23}\text{mol}^{-1})(6.626\times10^{-34}\text{Js})(3\times10^8\text{ms}^{-1})}{242\times10^{-9}\text{m}}$
$=4.947 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$
$=494.7 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}$
$=494 \mathrm{~kJ}\mathrm{~mol}^{-1}$
View full question & answer→Question 143 Marks
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength $6800$ $\mathring{\text{A}}.$ Calculate threshold frequency ($ν_0 $) and work function ($W_0$ ) of the metal.
AnswerThreshold wavelength of radian $(\lambda_0)=6800\mathring{\text{A}}=6800\times10^{-10}\text{m}$
Threshold frequency ($v_0$) of the metal
$=\frac{\text{c}}{\lambda_0}=\frac{3\times10^8\text{ms}^{-1}}{6.8\times10^{-7}\text{m}}=4.41\times10^{14}\text{s}^{-1}$
Thus, the threshold frequency ($v_0$) of the metal is $4.41 \times 10^{14} s^{–1}.$
Hence, work function ($W_0$) of the metal $= hν_0$
$= (6.626 \times 10^{–34}Js)(4.41 \times 10^{14} s^{–1})$
$= 2.922 \times 10^{–19}J$
View full question & answer→Question 153 Marks
Find:
- The total number and.
- The total mass of neutrons in 7 mg of 14C.
(Assume that mass of a neutron = 1.675 × 10–27kg).
AnswerNumber of atoms in ${ }^{14} C$ in $1 mole=6.022 \times 10^{23}$ atoms
1 atom of ${ }^{14} C$ contains $=14-6=8$ neutrons.
$\therefore$ The number of neutrons in 14 g of ${ }^{14} C =6.022 \times 10^{23} \times 8$ neutrons
Number of neutrons in $7 mg =\frac{\left(6.022 \times 10^{23} \times 8 \times 7\right)}{14000}=2.4088 \times 10^{21}$ netrons
Mass of one neutron $=1.674 \times 10^{-27} kg$
Mass of total neutrons in 7 g of ${ }^{14} C =\left(2.4088 \times 10^{21}\right)\left(1.675 \times 10^{-27} kg\right)=4.035 \times 10^{-6} kg$
View full question & answer→Question 163 Marks
Find energy of each of the photons which correspond to light of frequency $3 \times 10^{15} \mathrm{~Hz}$.
AnswerEnergy ( E ) of a photon is given by the expression,
$E=h v$
Where,
$\mathrm{h}=$ Planck's constant $=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
$v=$ frequency of light $=3 \times 10^{15} \mathrm{~Hz}$
Substituting the values in the given expression of E:
$E=\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{15}\right)$
$E=1.988 \times 10^{-18} \mathrm{~J}$
View full question & answer→Question 173 Marks
A certain particle carries $2.5 \times 10^{–16}C$ of static electric charge. Calculate the number of electrons present in it.
AnswerCharge on one electron $=1.6022 \times 10^{-19} \mathrm{C}$
$\Rightarrow 1.6022 \times 10^{-19} \mathrm{C}$ charge is carried by 1 electron.
$\therefore$ Number of electrons carrying a charge of $2.5 \times 10^{-16} \mathrm{C}$
$=\frac{1}{1.6022\times10^{-19}\text{C}}(2.5\times10^{-16}\text{C})$
$=1.560\times10^{3}\text{C}$
$=1560\text{C}$
View full question & answer→Question 183 Marks
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?
Answer Nuclear charge experienced by an electron (present in a multi-electron atom) is dependant upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge also decreases.
Among p-orbitals, 4p orbitals are farthest from the nucleus of bromine atom with (+35) charge. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the 2p and 3p orbitals along with the s-orbitals. Therefore, they will experience the lowest nuclear charge.
View full question & answer→Question 193 Marks
Symbols $\text{ }^{79}_{35}\text{Br}$ and $\text{ }^{79}\text{Br}$ can be written, whereas symbols $\text{ }^{35}_{79}\text{Br}$ and $\text{ }^{35}\text{Br}$ are not acceptable. Answer briefly.
AnswerThe general convention of representing an element along with its atomic mass (A) and atomic number (Z) is $\text{ }^{\text{A}}_{\text{Z}}\text{X}.$
Hence, $\text{ }^{79}_{35}\text{Br}$ is acceptable but $\text{ }^{35}_{79}\text{Br}$ is not acceptable.
$\text{ }^{79}\text{Br}$ can be written but $\text{ }^{35}\text{Br}$ cannot be written because the atomic number of an element is constant, but the atomic mass of an element depends upon the relative abundance of its isotopes. Hence, it is necessary to mention the atomic mass of an element.
View full question & answer→Question 203 Marks
Calculate the number of electrons which will together weigh one gram.
AnswerMass of one electron $=9.10939 \times 10^{-31} \mathrm{~kg}$
$\therefore$ Number of electrons that weigh $9.10939 \times 10^{-31} \mathrm{~kg}=1$
Number of electrons that will weigh $1 \mathrm{g}=\left(1 \times 10^{-3} \mathrm{~kg}\right)$
$=\frac{1}{9.10939\times10^{-31}\text{kg}}\times(1\times10^{-3}\text{kg})$
$=0.1098\times10^{-3}+31$
$=0.1098\times10^{28}$
$=1.098\times10^{27}$
View full question & answer→Question 213 Marks
What is the number of photons of light with a wavelength of $4000$ pm that provide $1J$ of energy?
AnswerEnergy $(E)$ of a photon $=h \nu$
Energy $\left(E_n\right)$ of ' $n$ ' photons $=n h v$
$\Rightarrow n=\frac{E_{n} \lambda}{hc}$
Where,
$\lambda=$ wavelength of light $=4000 pm =4000 \times 10^{-12} m$
$c=$ velocity of light in vacuum $=3 \times 10^8 m / s$
$h =$ Planck's constant $=6.626 \times 10^{-34} Js$
Substituting the values in the given expression of $n$ :
$n=\frac{(1) \times\left(4000 \times 10^{-12}\right)}{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^8\right)}=2.012 \times 10^{16}$
Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are $2.012 \times 10^{16}$.
View full question & answer→Question 223 Marks
If the diameter of a carbon atom is 0.15nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20cm long.
Answer$1 \mathrm{~m}=100 \mathrm{~cm}$
$1 \mathrm{~cm}=10^{-2} \mathrm{~m}$
Length of the scale $=20 \mathrm{~cm}$
$=20 \times 10^{-2} \mathrm{~m}$
Diameter of a carbon atom $=0.15 \mathrm{~nm}$
$=0.15 \times 10^{-9} \mathrm{~m}$
One carbon atom occupies $0.15 \times 10^{-9} \mathrm{~m}$.
$\therefore$ Number of carbon atoms that can be placed in a straight line
$=\frac{20\times10^{-2}\text{m}}{0.15\times10^{-9}\text{m}}$
$=133.33\times10^7$
$=1.33\times10^9$
View full question & answer→Question 233 Marks
A 25watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.
AnswerEnergy emitted by the bulb = 25watt = $25Js^{-1}$
Energy of one photon (E) $=\text{hv}=\frac{\text{hc}}{\lambda}$
$\lambda=0.57\mu\text{m}=0.57\times10^{-6}\text{m}$
$\text{E}=\frac{(6.626\times10^{-34}\text{Js}\times3.0\times10^8\text{ms}^{-1})}{0.57\times10^{-6}\text{m}=3.48\times10^{-19}\text{J}}$
$\therefore$ No. of photons emitted per sec $=\frac{25\text{Js}^{-1}}{3.48\times10^{-19}\text{J}=7.18\times10^{19}}$
View full question & answer→Question 243 Marks
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800pm, calculate the characteristic velocity associated with the neutron.
AnswerFrom de Broglie’s equation,
$\lambda=\frac{\text{h}}{\text{mv}}$
$\text{v}=\frac{\text{h}}{\text{m}\lambda}$
Where,
v = velocity of particle (neutron)
h = Planck’s constant
m = mass of particle (neutron)
$\lambda$ = wavelength
Substituting the values in the expression of velocity (v),
$\text{v}=\frac{6.626\times10^{-34}\text{Js}}{(1.67493\times10^{-27}\text{kg})(800\times10^{-12}\text{m})}$
$=4.94\times10^2\text{ms}^{-1}$
$\text{v}=494\text{ms}^{-1}$
$\therefore$ Velocity associated with the neutron = $494 \mathrm{ms}^{-1}$
View full question & answer→Question 253 Marks
Write down the quantum numbers n, and I for the following orbitals.
i. $3 d_{x^2-y^2}$
ii. $4 \mathrm{~d}_{\mathrm{z}^2}$
iii. $3 \mathrm{~d}_{\mathrm{xy}}$
iv. $4 d_{x z}$
v. $2 p_z$
vi. $3 p_x$
Answer
- n = 3, l = 2,
- n = 4, l = 2
- n = 3, l = 2
- n = 4, l = 2,
- n = 2, l = 1,
- n = 3, l = 1.
View full question & answer→Question 263 Marks
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/ have the same energy lists:
- $\text{n}=4,\text{l}=2,\text{m}_\text{l}=-2,\text{m}_{\text{s}}=-\frac{1}{2}$
- $\text{n}=3,\text{l}=2,\text{m}_\text{l}=1,\text{m}_{\text{s}}=+\frac{1}{2}$
- $\text{n}=4,\text{l}=1,\text{m}_\text{l}=0,\text{m}_{\text{s}}=+\frac{1}{2}$
- $\text{n}=3,\text{l}=2,\text{m}_\text{l}=-2,\text{m}_{\text{s}}=-\frac{1}{2}$
- $\text{n}=3,\text{l}=1,\text{m}_\text{l}=-1,\text{m}_{\text{s}}=+\frac{1}{2}$
- $\text{n}=4,\text{l}=1,\text{m}_\text{l}=0,\text{m}_{\text{s}}=+\frac{1}{2}$
Answer For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.
Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1(4d).
View full question & answer→Question 273 Marks
- Which orbital is non-directional?
- What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to n = 2 state in the hydrogen atom?
($h = 6.626 \times 10^{-34}Js$)Answer
- s-Orbital is spherically symmetrical, i.e. it is non directional.
- Transition is from $n_1 = 5$ to $n_2 = 2$ state,
$\Delta\text{E}=2.18\times10^{-18}\text{J}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$
$=2.18\times10^{-18}\text{J}\Big(\frac{1}{5 ^2}-\frac{1}{3^2}\Big)$
$=-4.58\times10^{-19}\text{J}$
The negative sign of energy means it is an emission energy.
Frequency $=\frac{\text{Energy}}{\text{Plant's constant}}$
$\Rightarrow\text{v}=\frac{\Delta\text{E}}{\text{h}}=\frac{4.58\times10^{19}\text{J}}{6.626\times10^{-34}\text{Js}}$
$=6.91\times10^{14}\text{Hz}$
Wavelength $=\frac{\text{Velocity}}{\text{Frequency}}$
$\Rightarrow\lambda=\frac{\text{c}}{\text{v}}=\frac{3\times10^8\text{ms}^{-1}}{6.91\times10^{14}\text{Hz}}$
$\lambda=0.434\times10^{-6}\text{m}$
$\lambda=434\times10^{-9}\text{m}=434\text{nm}$ $[\because1\text{nm}=10^{-9}\text{m}]$ View full question & answer→Question 283 Marks
- The frequency of the strong yellow line in the spectrum of sodium is $5.09 \times 10^{14} s^{-1}$. Calculate the wavelength of the light in nanometer.
- Using s, p, d notations, describes the orbital with the following quantum numbers:
- $n = 3, l = 1, m = 0, (b) n = 1, l = 0$
- Which quantum number distinguishes the electron in the same orbital? Name the principle involved.
Answer
- Given, $\text{n}=5.09\times10^{14}\text{s}^{-1},$
$\text{c}=3\times10^8\text{ms}^{-1},\lambda=\ ?$
Wavelenth, $\lambda=\frac{\text{c}}{\text{v}}=\frac{3\times10^8\text{ms}^{-1}}{5.09\times10^{14}\text{s}^{-1}}$
$=5.89\times10^{-7}\text{m}$
$=5.89\times10^{-7}\times10^9\text{nm}=589\text{nm}$
- (a) $3p_y$; (b) 1s
- Spin quantum number, Pauli exclusion principle.
View full question & answer→Question 293 Marks
Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.
Answer$\bar{\text{v}}=\text{R}_\text{H}\Big[\frac{1}{\text{n}_1^2}-\frac{1}{\text{n}_2^2}\Big]$
$\bar{\text{v}}=109677\text{cm}^{-1}\Big[\frac{1}{2^2}-\frac{1}{3^2}\Big]$
$\Rightarrow\bar{\text{v}}=109677\times\frac{5}{36}=1523.9\text{cm}^{-1}$
$\bar{\text{v}}=\frac{1}{\lambda}\Rightarrow\lambda=\frac{1}{\bar{\text{v}}}=\frac{1}{1523.9}=6564\times10^{-5}\text{cm}$
$\lambda=6564\times10^{-7}\text{m}$
$\text{E}=\frac{\text{hc}}{\lambda}=\frac{6.626\times10^{-34}\times3\times10^8}{6564\times10^{-7}}0$
$=3.028\times10^{-19}\text{J}$
View full question & answer→Question 303 Marks
Calculate the wavelength of a 1000kg rocket moving with a velocity of 3000km/ hour.
Answer$\text{m}=1000\text{kg}$
$\text{c}=3000\text{kg/ hour}$
$=\frac{3000\times1000}{60\times60}\text{ms}^{-1}$
Using de Brogloie equation,
$\lambda=\frac{\text{h}}{\text{mc}}=\frac{6.626\times10^{-34}\times60\times60}{1000\text{kg}\times3000\times1000}$
$\lambda=\frac{6.626\times36}{3}\times10^{-34+2-9}$
$\lambda=79.512\times10^{-41}\text{m}$
$=7.9512\times10^{-40}\text{m}$
View full question & answer→Question 313 Marks
What is the difference between the terms orbit and orbital?
Answer
|
|
Orbit
|
|
Orbital
|
|
i.
|
An orbit is a well-defined circular path around the nucleus in which the electrons revolve.
|
i.
|
An orbital is the three-dimensional space around the nucleus within which the probability of finding an electron is maximum (upto 90%)
|
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ii.
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It represents the planar motion of an electron around the nucleus.
|
ii.
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It represents the three dimensional motion of an electron around the nucleus.
|
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iii.
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All orbits are circular and disc like.
|
iii.
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Different orbitals have different shapes, i.e. s-irbitals are spherically symmetrical, p-orbitals are dumb-bell shaped and so on.
|
View full question & answer→Question 323 Marks
- The mass of an electron is $9.1 \times 10^{-28}g$. If its K.E. is $3.0 \times 10^{-25}J$, calculate its wave-length in Angstrom. $[h = 6.6 \times 10^{-34}Js]$
- What is photoelectric effect?
Answer
- $\text{m}=9.1\times10^{-28}\text{g}=9.1\times10^{-31}\text{kg}$
$\text{K.E}=3.0\times10^{-25}\text{J}$
$\frac{1}{2}\text{mv}^2=3.0\times10^{-25}\text{J}$
$\text{K.E}=\frac{1}{2}\text{mv}^2$
$\Rightarrow\text{V}=\sqrt{\frac{2\text{KE}}{\text{m}}}$
$\lambda=\frac{\text{h}}{\text{mV}}=\frac{\text{h}}{\text{m}\times\sqrt{\frac{2\text{K.E}}{\text{m}}}}=\frac{\text{h}}{\sqrt{2\times\text{K.E}\times\text{m}}}$
$\lambda=\frac{\text{h}}{\sqrt{2\text{m}\times\text{K.E}}}=\frac{6.6\times10^{-34}\text{Js}}{\sqrt{2\times9.1\times10^{-31}\times3\times10^{-25}}}$
$\lambda=\frac{6.6\times10^{-34}}{\sqrt{54.6\times10^{-56}}}=\frac{6.6\times10^{-34}}{7.39\times10^{-28}}$
$=8.93\times10^{-7}\text{m}=893\text{nm}=8930\mathring{\text{A}}$
- When a beam of light having frequency more than threshold frequency is made to fall on metals like alkali metals, electrons are ejected. These electrons are called photoelectrons and this phenomenon is called photoelectric effect.
View full question & answer→Question 333 Marks
According to de Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100g does not move like a wave when it is thrown by a bowler at a speed of 100km/ h. Calculate the wavelength of the ball and explain why it does not show wave nature.
Answer de Broglie equation,
$\lambda=\frac{\text{h}}{\text{mc}},\text{m}=100\text{g}=\frac{100}{1000}=0.1\text{kg},$
Velocity of cricket ball c,
$=100\text{km/ h}$
$=\frac{100\times1000\text{m}}{60\times60\text{s}}=-\frac{1000}{36}\text{ms}^{-1},$
Planck's constant, $\text{h}=6.626\times10^{-34}\text{Js}$
Wavelength,
$\lambda=\frac{6.626\times10^{-34}\text{Js}}{0.1\text{kg}\times\frac{1000}{36}\text{ms}^{-1}}=6.626\times10^{-36}\times36\text{m}$
$\lambda=2.385\times10^{-34}\text{m}$
Since mass of the ball is large, therefore, $'\lambda'$ is very small, the wave nature cannot be observed.
View full question & answer→Question 343 Marks
Table-tennis ball has a mass 10g and a speed of 90m/s. If speed can be measured within an accuracy of 4% what will be the uncertainty in speed and position?
Answer Uncertainty in the speed of ball
$=\frac{90\times4}{100}=\frac{360}{100}=3.6\text{ms}^{-1}$
Uncertainty in position $=\frac{\text{h}}{4\pi\text{m}\Delta\text{v}}$
$=\frac{6.626\times10^{-34}\text{Js}}{4\times3.14\times10\times10^{-3}\text{kg g}^{-1}\times3.6\text{ms}^{-1}}$
$=1.46\times10^{-33}\text{m}$
View full question & answer→Question 353 Marks
The arrangement of orbitals on the basis of energy is based upon their (n + l) value. Lower the value of (n + l), lower is the energy. For orbitals having same values of (n + l), the orbital with lower value of n will have lower energy.
- Based upon the above information, arrange the following orbitals in the increasing order of energy.
- 1s, 2s, 3s, 2p
- 4s, 3s, 3p, 4d
- 5p, 4d, 5d, 4f, 6s
- 5f, 6d, 7s, 7p
- Based upon the above information, solve the questions given below:
- Which of the following orbitals has the lowest energy?
4d, 4f, 5s, 5p
- Which of the following orbitals has the highest energy?
5p, 5d, 5f, 6s, 6p
Answer
-
- (n + l) values are 1s = 1 + 0 = 1, 2s = 2 + 0 = 2, 3s = 3 + 0 = 3, 2p = 2 + 1 = 3
Hence, increasing order of their energy is 1s < 2s < 2p < 3s.
- 4s = 4 + 0 = 4, 3s = 3 + 0 = 3, 3p = 3 + 1 = 4, 4d = 4 + 2 = 6.
Hence, 3s < 3p < 4s < 4d.
- 5p = 5 + 1 = 6, 4d = 4 + 2 = 6, 5d = 5 + 2 = 7, 4f = 4 + 3 = 7, 6s = 6 + 0 = 6.
Hence, 4d < 5p < 6s < 4f < 5d.
- 5f = 5 + 3 = 8, 6d = 6 + 2 = 8, 7s = 7 + 0 = 7, 7p = 7 + 1 = 8.
Hence, 7s < 5f < 6d < 7p.
-
- 4d = 4 + 2 = 6, 4f = 4 + 3 = 7, 5s = 5 + 0 = 5, 7p = 7 + 1 = 8.
Hence, 5s has the lowest energy.
- 5p = 5 + 1 = 6, 5d = 5 + 2 = 7, 5f = 5 + 3 = 8, 6s = 6 + 0 = 6, 6p = 6 + 1 = 7.
Hence, 5f has the highest energy.
View full question & answer→Question 363 Marks
- List two main differences between orbit and orbital.
- If an electron is moving with a velocity 600m/ s which is accurate upto 0.005%, then calculate the uncertainty in its position.
($h = 6.626 \times 10^{-34}Js$ and mass of election = $9.11 \times 10^{-31}kg$)Answer
- Differences between orbit and orbitals:
| S. No |
Orbit
|
Orbital
|
| 1. |
Orbit is a well defined2-D circular path around the nucleus in which the electrons revolve.
|
Orbital is a 3-D space around the nucleus within which the probability of finding the electrons is maximum
|
| 2. |
Concept of orbit is not in Accor-dance with the wave nature of electrons.
|
It is in accordance with the wave nature of electrons.
|
| 3. |
Orbits do not have directional characteristics.
|
All orbital’s except s-orbital have directional characteristics.
|
- Uncertainty in speed, $\Delta\text{V}=\frac{0.005}{100}\times600\text{m/ s}$
$=0.03\text{ms}^{-1}$
Heisenberg Uncertainty Principle,
$\Delta\text{x} \times\text{m}\Delta\text{V}=\frac{\text{h}}{4\pi},\Delta\text{x}=$ Uncertainty in position,
$\Rightarrow\Delta\text{x}=\frac{6.626\times10^{-34}\text{Js}}{4\times\frac{22}{7}\times9.11\times10^{-31}\text{kg}\times0.03\text{ms}^{-1}}$
$=1.93\times10^{-3}\text{m}$ View full question & answer→Question 373 Marks
Hydrogen atom has only one electron, so mutual repulsion between electrons is absent. However, in multielectron atoms mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?
Answer The energy of an electron in a hydrogen atom is determined solely by the principal quantum number. Thus, the energy of the orbitals increases as follows:
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < (2.23),
The energy of an electron in a multielectron atom, that of the hydrogen atom, depends not only on its principal quantum number (shell), but also on its azimuthal quantum number (subshell). That is, for a given principal quantum number, s, p, d, f .... all have different energies.
View full question & answer→Question 383 Marks
- What is the shape of orbital with l = 2 and l = 3?
- Account for the following:
- Chromium has configuration $3d^5 4s^1$ and not $3d^4 4s^2$.
- Bohr's orbits are called stationary orbits or states.
Answer
- l = 2, i.e. d-orbital has double dumb-bell shape, l = 3. i.e. f-orbital, having 7orbitals: four of them are triple dumb-bells in a hexagonal pattern, two are quadruple dumb-bells and one is a dumb-bell with a double donut.
-
- It is due to stability of half filled orbitals.
- It is because electrons do not radiate energy as long as they remain in the same energy level.
View full question & answer→Question 393 Marks
A bulb emits light of wavelength $4500\mathring{\text{A}}.$ The bulb is rated as 150 watt and 8% energy is emitted as light. How many photons are emitted by the bulb per second?
$\left[\mathrm{h}=6.626 \times 10^{-34} \mathrm{J} \mathrm{s}\right]$
AnswerEnergy of 1 photon $=\frac{\text{hc}}{\lambda}$
$=\frac{6.626\times10^{-34}\text{Js}\times3\times10^8\text{ms}^{-1}}{4500\times10^{-10}\text{m}}$
$=\frac{19.878}{45\times100}\times10^{-34+8+10}\text{J}$
$=\frac{19.878}{45}\times10^{-16-2}\text{J}=\frac{19.878}{45}\times10^{-18}\text{J}$
$=\frac{198.78}{45}\times10^{-19}\text{J}=4.417\times10^{-19}\text{J}$
Power of bulb $=150 \mathrm{watt} =150 \mathrm{Js}^{-1}$
Energy radiated as light per second,
$=150\text{Js}^{-1}\times\frac{8}{100}=12\text{Js}^{-1}$
Number of photons emitted by bulb per second,
$=\frac{\text{Total energy radiate per second}}{\text{Energy of 1 photon}}$
$=\frac{12\text{Js}^{-1}}{4.417\times10^{-19}\text{J}}=2.716\times10^{19}$ photons per second
View full question & answer→Question 403 Marks
i. What is de-Broglie wavelength of an electron moving with velocity of light? Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$, $\mathrm{h}=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}$.
ii. What is angular momentum of electron in the $5^{\text {th }}$ shell?
Answer
- $\text{m}=9.1\times10^{-31}\text{kg},$
$\text{c} =3.0\times10^8\text{ms}^{-1}$
$\lambda=\ ?,\text{h}=6.626\times10^{-34}\text{Js}$ or $\text{Kg}/\ \text{m}^2\text{s}^{-1}$
Applying de Broglie equation:
$\lambda=\frac{\text{h}}{\text{mc}}=\frac{6.626\times10^{-34}\text{kg/ m}^2\text{s}^{-1}}{9.1\times10^{-31}\text{kg}\times3.0\times10^8\text{ms}^{-1}}$
$\lambda=\frac{6.626}{27.3}\times10^{-34+31-8}\text{m}$
$\lambda=\frac{6.626}{27.3}\times10^{-11}=\frac{66.26}{27.3}\times10^{-12}\text{m}$
$\lambda=2.427\times10^{-12}\text{m}$
- Angular momentum mvr $=\frac{\text{nh}}{2\pi}$
For $5^{\text {th }}$ energy level, $\text{n}=5$
$\therefore\text{mvr}=\frac{5\text{h}}{2\pi}$ View full question & answer→Question 413 Marks
According to de Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100g does not move like a wave when it is thrown by a bowler at a speed of 100km/h. Calculate the wavelength of the ball and explain why it does not show wave nature.
Answer $\lambda=\frac{\text{h}}{\text{mv}}$
$\text{m}=100\text{g}=0.1\text{kg.}$
$\text{v}=100\text{km/hr}=\frac{100\times1000\text{m}}{60\times60\text{s}}=\frac{1000}{36}\text{ms}^{-1}$
$\text{h}=6.626\times10^{-34}\text{Js}$
$\lambda=\frac{6.626\times10^{-34}\text{Js}}{0.1\text{kg}\times\frac{1000}{36}\text{ms}^{-1}}=6.626\times10^{-36}\times36\text{m}^{-1}$
$=238.5\times10^{-36}\text{m}^{-1}$
View full question & answer→Question 423 Marks
Calculate the wavelength of the spectral line obtained in the spectrum of $Li^{2+}$ ion when the transition takes place between two levels whose sum is 4 and the difference is 2.
AnswerSuppose the transition takes place between levels $n_1$ and $n_2$
Then, $n + n_2 = 4$ and $n_2 – n_1 = 2$
By solving these equations, we get $n_1 = 1, n_2 = 3$,
$\therefore\frac{1}{\lambda}=\text{R}\Big(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Big)\text{Z}^2$
For $Li^{2+}$, Z = 3
$\therefore\frac{1}{\lambda}=109677\text{cm}^{-1}\Big(\frac{1}{1^2}-\frac{1}{3^2}\Big)\times3^2$
$=109677\times\Big(\frac{1}{1}-\frac{1}{9}\Big)\times9\text{cm}^{-1}$
$=109677\times8\text{cm}^{-1}$
or $\lambda=\frac{1}{109677\times8\text{cm}^{-1}}$
$=1.14\times10^{-6}\text{cm}$
View full question & answer→Question 433 Marks
When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of $1.68 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photo-electron to be emitted?
$\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)$
AnswerThe energy (E) of a 300nm photon is given by Planck's quantum theory.
$\text{E}=\text{hv}=\frac{6.626\times10^{34}\text{Js}\times3.010^8\text{ms}^{-1}}{300\times10^{-9}\text{m}}$
$\text{E}=6.626\times10^{-19}\text{J}$
The energy of 1 mole of photons,
$=6.626\times10^{-19}\text{J}\times6.022\times10^{23}\text{mol}^{-1}$
$=3.99\times10^5\text{J/ mol}^{-1}$
Energy of incident light = Threshold energy + kinetic energy,
$\therefore$ Threshold energy = Energy of incident light - kinetic energy,
$=(3.99\times10^5-1.68\times10^5)\text{J/ mol}^{-1}$
$=2.31\times10^5\text{J/ mol}^{-1}$
Threshold energy is the minimum energy needed to remove a mole of electrons from sodium,
$=(3.99-168)10^5\text{J/ mol}^{-1}$
$=2.31\times10^5\text{J/ mol}^{-1}$
The minimum energy for one electron,
$=\frac{2.31\times10^5\text{J/ mol}^{-1}}{6.022\times10^{23}\text{electrons mol}^{-1}}$
$=\frac{\text{Energy per mol}^{-1}}{\text{Avogadro's number}}$
$=3.84\times10^{-19}\text{J/ electron}$
This corresponds to the wavelength calculated as follows:
$\lambda=\frac{\text{hc}}{\text{E}}=\frac{6.626\times10^{-34}\text{Js}\times3.0\times10^8\text{ms}^{-1}}{3.84\times10^{-19}\text{J}}$
$=5.17\times10^{-7}\text{m}$
$\lambda=5.17\times10^{-7}\times10^9\text{nm}$
$\Rightarrow\lambda=517\text{nm}$
(This corresponds to green light).
View full question & answer→Question 443 Marks
Which of the following sets of orbitals are degenerate and why?
i. $1 \mathrm{~s}, 2 \mathrm{~s}$ and 3 s in Mg -atom.
ii. $2 p_x, 2 p_y$ and $2 p_z$ in C -atom.
iii. $3 \mathrm{~s}, 3 \mathrm{p}_{\mathrm{x}}$ and 3 d -orbitals in H -atom.
Answeri. $1 \mathrm{~s}, 2 \mathrm{~s}$ and 3 s -orbitals in Mg -atom are not degenerate because these have different values of n .
ii. $2 p_x, 2 p_y$ and $2 p z$-orbitals in C -atom are degenerate because these belong to same subshell.
iii. $3 \mathrm{~s}, 3 \mathrm{p}_{\mathrm{x}}$ and 3 d -orbitals in H -atom are degenerate because for H -atom, the subshells having same value of n have same energies.
View full question & answer→Question 453 Marks
What are the frequency and wavelength of a photon during a transition from $\mathrm{n}=5$ state to $\mathrm{n}=2$ state in the $\mathrm{He}^{+}$ion.
Answer$\text{n}_1=2$
$\text{n}_2=5$
By Rydberg equation
$\frac{1}{\lambda}=\text{R}_{\text{H}}\text{Z}^2\Big[\frac{1}{\text{n}_1{^2}}-\frac{1}{\text{n}_2{^2}}\Big]$ $\text{R}_{\text{H}}=1.09\times10^7\text{m}^{-1}\\ \text{Z}=2$
$=1.09\times10^7\times4\Big[\frac{1}{2^2}-\frac{1}{5^2}\Big]$
$=4.36\times10^7\Big[\frac{25-4}{25\times4}\Big]$
$=4.36\times10^{-5}\times21$
$\lambda=\frac{1}{21\times4.36\times10^{-5}}$
$=0.01\times10^{-5}\text{m}$
$\nu=\frac{\text{c}}{\lambda}=\frac{3\times10^8}{0.01\times10^{-5}}$
$=300\times10^{13}\text{m}^{-1}$
View full question & answer→Question 463 Marks
- Define Aufbau's principle.
- Draw the shape of $\text{d}_{\text{z}^2}$ orbital.
- Which quantum number specify number of orbitals in a given subshell?
Answer
- Aufbau's principle: Electrons are filled in the orbitals in increasing order of energies in grand state.
-
- Number of orbitals in gives subshell is decided by magnetic quantum number.
View full question & answer→Question 473 Marks
The first shell may contain up to 2 electrons, the second shell up to 8, the third shell up to 18, and the fourth shell up to 32. Explain this arrangement in terms of quantum numbers.
Answer For the fust shell $\text{n}=1,\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2}$ and $-\frac{1}{2}.$ It can have 2 electrons both with opposite spins.
- For $\text{n}=2,$
$\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2},$
$\text{l}=1,\text{m}_\text{l}=1,0+1,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}.$
Therefore, a total of 2 + 6 = 8 electrons are present.
- For $\text{n}=3,$ when
$\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$
$\text{l}=1,\text{m}_\text{l}=-1,0+1,\text{m}_\text{s}=\frac{1}{2},-\frac{1}{2}$
$\text{l}=2,\text{m}_\text{l}=-2,-1,0,+1,+2,,\text{m}_\text{s}$
$=+\frac{1}{2},-\frac{1}{2}$
Therefore, a total of 2 + 6 + 10 = 18 electlons aie present.
- For $\text{n}=4,$ when
$\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$
$\text{l}=1,\text{m}_\text{l}=-1,0+1,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$
$\text{l}=3,\text{m}_\text{l}=-2,-1,0+1,+2,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$
$\text{l}=3,\text{m}_\text{l}=-3,-2,-1,0,+1,+2,+3,\text{m}_\text{s}$
$=+\frac{1}{2},-\frac{1}{2}$
Therefore, a total of 2 + 6 + 10 + 14 = 32 electrons are present. View full question & answer→Question 483 Marks
- What is the common between $d_{xy}$ and $d_{x^2 - y^2}$ orbitals?
- What is the difference between them?
- What is the angle between the lobes of the above two orbitals?
Answer
- Both have identical shape, consisting of four lobes.
- Lobes of $d_{x^2 - y^2}$ lie along the x and y-axes whereas those of $d_{xy}$ lie between x and y-axes.
- 45°
View full question & answer→Question 493 Marks
- An orbital has n = 2, what are possible values of I and $m_l$?
- List the quantum numbers $m_l$ and I of electrons for 3d-orbital?
- Which of the following orbitals are possible?
2d, 1s, 2p, 3fAnswer
- n = 2, l = 0, m = 0
n = 2, l = 1, m = -1, 0, + 1
- l = 2, $m_l$ = -2, - 1, 0, + 1, + 2
- 1s and 2p orbitals are possible.
View full question & answer→Question 503 Marks
What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is $-2.18 \times 10^{-11}$ erg.
Answer$\Delta\text{E}=\text{E}_5-\text{E}_1$
$=2.18\times10^{-11}\Big(\frac{1}{\text{n}^2_\text{i}}-\frac{1}{\text{n}^2_\text{f}}\Big)\text{erg}$ $[\text{n}_\text{i}=1$ and $\text{n}_\text{f}=5]$
$\Delta\text{E}=2.18\times10^{-11}\Big(\frac{1}{1^2}-\frac{1}{5^2}\Big)\text{erg}$
$\Delta\text{E}=2.18\times10^{-11}\times\frac{24}{25}$
$=2.0928\times10^{-11}\text{erg}$
$=2.0928\times10^{-18}\text{J}$ $[\because1\text{erg}=10^{-7}\text{J}]$
When electron retums to ground state, it emits enrrgy equals to $\Delta\text{E}$ hence,
$\Delta\text{E}=\frac{\text{hc}}{\lambda}$
or $\lambda=\frac{\text{hc}}{\Delta\text{E}}=\frac{6.626\times10^{-34}\text{Js}\times3.0\times10^8\text{ms}^{-1}}{2.0928\times10^{-18}\text{J}}$
$=9.498\times10^{-8}\text{m}$
$=949.8\times10^{-10}\text{m}$
$\text{m}=949.8\mathring{\text{A}}$
View full question & answer→Question 513 Marks
What is the experimental evidence in support of the idea that electronic energies in an atom are quantized?
Answer The bright line spectrum shows that the energy levels in an atom are quantized. These lines corresponds to definite wavelength sand are obtained as a result of electronic transitions between the energy levels. Hence, the electrons in these levels have quantized values.
View full question & answer→Question 523 Marks
- Define principal quantum number(n).
- Write the electronic configuration of $Cr^+$ [Atomic number of Cr = 24].
- Define Pauli's exclusion principle.
Answer
- Principal quantum number tells the principal energy level or shell to which the electron belongs. It gives the information about the distance and the energy of the electron.
- $Cr^+ = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5$
- Pauli's exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers.
View full question & answer→Question 533 Marks
Find out the number of wave made by a Bohr electron in one complete revolution in its 3rd orbit.
Answer In general, number of waves in any orbit is,
Number of wayes $=\frac{\text{Circumference of orbit}}{\text{Wavelength}}=\frac{2\pi\text{r}}{\lambda}$
But $\lambda=\frac{\text{h}}{\text{mc}}$
Number of waves $=\frac{2\pi\text{r}}{\frac{\text{h}}{\text{mv}}}=\frac{2\pi\text{r}.\text{mv}}{\text{h}}=\frac{2\pi(\text{mvr})}{\text{h}}$
The angular momentum of Bohr's 3rd orbit is,
$\text{mvr}=\frac{3\text{h}}{2\pi}$
$\therefore$ Number of wayes $=\frac{2\pi}{\text{h}}\times\frac{3\text{h}}{2\pi}=3$
Number of waves in Bohr,s 3rd orbit = 3.
View full question & answer→Question 543 Marks
The radius of first Bohr orbit of hydrogen atom is $0.529\mathring{\text{A}}.$ Calculate the radii of:
- The third orbit of $He^+$ ion,
- The second orbit of $Li^2+$ ion.
AnswerRadius of nth Bohr orbit, $\text{r}_\text{n}=\frac{\text{n}^2\text{h}^2}{4\pi^2\text{m}.\text{Ze}^2}$
For hydrogen atom $Z = 1$, first orbit $n = 1$
$\text{r}_1=\frac{\text{h}^2}{4\pi^2\text{me}^2}=0.529\mathring{\text{A}}$
- For $He^+$ ion, $Z = 2,$ third orbit, $n = 3$
$\text{r}_3(\text{He}^+)=\frac{3^2\text{h}^2}{4\pi^2\text{m}\times2\times\text{e}^2}$
$=\frac{9}{2}\Big[\frac{\text{h}^2}{4\pi^2\text{me}^2}\Big]=\frac{9}{2}\times0.529=2.380\mathring{\text{A}}$
- For $Li^{2+}$ ion, $Z = 3$, second orbit, $n = 2$
$\text{r}_2(\text{Li}^{2+})=\frac{2^2\text{h}^2}{4\pi^2\text{m}\times3\times\text{e}^2}=\frac{4}{3}\Big[\frac{\text{h}^2}{4\pi^2\text{me}^2}\Big]$
$=\frac{4}{3}\times0.529=0.7053\mathring{\text{A}}$ View full question & answer→Question 553 Marks
Calculate the uncertainty in the position of a dust particle with mass equal to 1 mg if the uncertainty in velocity is 5.5 $\times 10^{-20} \mathrm{~ms}^{-1} .\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{~Js}\right)$
Answer$\Delta \mathrm{v}=$ uncertainty in position = ?
$\Delta \mathrm{v}=$ Uncertainty in velocity $=5.5 \times 10^{-20} \mathrm{~ms}^{-1}$
$\text{m}=1\text{mg}=10^{-6}\text{kg}$ $[\because1\text{ms}=10^{-6}\text{kg}]$
Applying Heisenberg's uncertainity principle,
$\Delta\text{x},\Delta\text{v}=\frac{\text{h}}{4\text{m}\pi}$
Uncretainity in position, $\Delta\text{x}=\frac{ \text{h}}{4\text{m}\pi\times\Delta\text{v}}$
$=\frac{6.626106^{-34}\text{Js}}{4\times10^{-6}\text{kg}\times3.14\times5.5\times10^{-20}\text{ms}^{-1}}\text{m}$
$\Delta\text{x}=\frac{6.626}{69.08}\times10^{-8}\text{m}$
$\Delta\text{x}=\frac{66.26}{69.08}\times10^{-9}\text{m}=0.959\times10^{-9} \ \text{m}$
$\Delta\text{r}=9.59\times10^{-10}\text{m}$
View full question & answer→Question 563 Marks
- What is the main difference between electro magnetic waves theory and Planck's quantum theory.
- Which rule is violated in the following orbital diagram:
Answer
- Electromagnetic Waves Theory: Energy is radiated or absorbed continuously.
Planck's Quantum Theory: Energy is radiated or absorbed not continuously but discontinuously in the form of small packets called quantas or photons.
- Hund's rule is being violated: Because orbitals having equal energy should be singly filled first and then pairing of electron should take place.
View full question & answer→Question 573 Marks
Calculate the wavelength of tennis ball of mass 60 gram moving with a velocity of $10 \mathrm{~ms}^{-1} .\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{Js}\right)$
AnswerAccording to de Broglie equation,
$\text{m}=60\text{g}=\frac{60}{1000}\text{kg}$
$\text{c}=10\text{ms}^{-1}$
$\text{h}=6.626\times10^{-34}\text{Js}$
$\lambda=\frac{\text{h}}{\text{mc}}$
$\lambda=\frac{6.626\times10^{-34}\text{Js}}{6\times10^{-2}\text{kg}\times10\text{ms}^{-1}}$
$\lambda=\frac{6.626}{6}\times10^{-33}\text{m}$
$\lambda=1.104\times10^{-33}\text{m}$
View full question & answer→Question 583 Marks
Correct the following electronic configuration of the elements in the ground state.
- $1\text{s}^2\ 2\text{s}^1,2\text{p}^2_\text{x},2\text{p}^2_\text{y},2\text{p}^2_\text{z},3\text{s}62,2\text{p}^1_\text{x}$
- $1\text{s}^2\ 2\text{s}^1,2\text{p}^1_\text{x},2\text{p}^1_\text{y},2\text{p}^1_\text{z}$
- $1\text{s}^2\ 2\text{s}^1,2\text{p}^6,3\text{s}^2,3\text{p}^6,3\text{d}^5$
- $1\text{s}^2\ 2\text{s}^2,2\text{p}^6,3\text{p}^6,3\text{d}^4,4\text{s}^2$
Answer
- $1\text{s}^2\ 2\text{s}^2,2\text{p}^2_\text{x},2\text{p}^2_\text{y},2\text{p}^2_\text{z},3\text{s}^2$
- $1\text{s}^2\ 2\text{s}^2,2\text{p}^1_\text{x},2\text{p}^1_\text{y},2\text{p}^1_\text{y},2\text{p}^1_\text{z}$
- $1\text{s}^2\ 2\text{s}^2,2\text{p}^6,3\text{s}^2,3 \text{p}^6,4\text{s}^2,3\text{d}^2$
- $1\text{s}^2\ 2\text{s}^2,2\text{p}^6,3\text{s} ^2,3\text{p}^6,3\text{d}^5,4\text{s}^1$
View full question & answer→Question 593 Marks
Chlorophyll present in green leaves of plants absorbs light at $4.620 \times 10^{14}Hz$. Calculate the wavelength of radiation in nanometer. Which part of the electromagnetic spectrum does it belong to?
AnswerFrequency and wavelength are related by the following equation:
$\mathrm{f}=\frac{\mathrm{c}}{\lambda}$ Where f is the frequency $=4.620 \times 10^{14} \mathrm{~Hz}$
$\mathrm{c}=$ speed of light $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
Therefore, $\lambda=\frac{\mathrm{c}}{\mathrm{f}}$
$=\frac{3 \times 10^8}{4.620 \times 10^{14}}$
$=6.49 \times 10^{-7} \mathrm{~m}=649 \mathrm{~nm}$
It belongs to the visible light of the spectrum.
View full question & answer→Question 603 Marks
Calculate the energy required for the process,
$\text{He}^+(\text{g})\rightarrow\text{He}^{2+}\text{(g)}+\text{e}^-$
The ionization energy for the H atom in the ground state is $2.18 \times 10^{-18} \mathrm{~J}\mathrm{~atom}^{-1}$
AnswerEnergy associated with hydrogen-like species is given by,
$\text{E}_{\text{n}}=-2.18\times10^{-18}\Big(\frac{\text{Z}^2}{\text{n}^2}\Big)\text{J}$
For ground state of hydrogen atom,
$\triangle\text{E = E}_{\infty}-\text{E}_1$
$=0-\bigg[-2.18\times10^{-18}\bigg\{\frac{(1)^2}{(1)^2}\bigg\}\bigg]\text{J}$
$\triangle\text{E}=2.18\times10^{-18}\text{J}$
For the given process,
$\text{He}^+(\text{g})\rightarrow\text{He}^{2+}\text{(g)}+\text{e}^-$
An electron is removed from $\text{n}=1\text{ to}\text{ n}=\infty.$
$\triangle\text{E = E}_{\infty}-\text{E}_1$
$=0-\bigg[-2.18\times10^{-18}\bigg\{\frac{(2)^2}{(1)^2}\bigg\}\bigg]$
$\triangle\text{E}=8.27\times10^{-18}\text{J}$
$\therefore$ The energy required for the process $8.27\times10^{-18}\text{J}.$
View full question & answer→Question 613 Marks
Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
AnswerFor the Balmer series, $\mathrm{n}_{\mathrm{i}}=2$. Thus, the expression of wave number $(\overline{\mathrm{v}})$ is given by,
$\overline{\mathrm{v}}=\left[\frac{1}{(2)^2}-\frac{1}{\mathrm{n}_{\mathrm{r}}^2}\right]\left(1.097 \times 10^7 \mathrm{~m}^{-1}\right)$
Wave number $(\overline{\mathbf{v}})$ is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, $\overline{\mathbf{v}}$ has to be the smallest.
For $\bar{v}$ to be minimum, $n_f$ should be minimum. For the Balmer series, a transition from $n_i=2$ to $n_f=3$ is allowed. Hence, taking $\mathrm{n}_{\mathrm{f}}=3$, we get:
$\bar{\text{v}}=(1.097\times10^7)\Big[\frac{1}{2^2}-\frac{1}{3^2}\Big]$
$\bar{\text{v}}=(1.097\times10^7)\Big[\frac{1}{4}-\frac{1}{9}\Big]$
$=(1.097\times10^7)\Big(\frac{9-4}{36}\Big)$
$=(1.097\times10^7)\Big(\frac{5}{36}\Big)$
$\bar{\text{v}}=1.5236\times10^6\text{ m}^{-1}$
View full question & answer→Question 623 Marks
The diameter of zinc atom is 2.6 $\mathring{\text{A}}$.Calculate,
- Radius of zinc atom in pm
- Number of atoms present in a length of 1.6cm if the zinc atoms are arranged side by side lengthwise.
Answer
- $\text{Radius of zinc atom }=\frac{\text{Diameter}}{2}$
$=\frac{2.6\mathring{\text{A}}}{2}$
$=1.3\times10^{-10}\text{m}$
$=130\times10^{-12}\text{m}=130\text{pm}$
- Length of the arrangement = 1.6cm
$=1.6 \times 10^{-2} \mathrm{~m}$
Diameter of zinc atom $=2.6 \times 10^{-10} \mathrm{~m}$
$\therefore$ Number of zinc atoms present in the arrangement
$=\frac{1.6\times10^{-2}\text{m}}{2.6\times10^{-10}\text{m}}$
$=0.6153\times10^8\text{m}$
$=6.153\times10^7$ View full question & answer→Question 633 Marks
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:
$\text{mvr}=\text{n}\frac{\text{h}}{2\pi}...(1)$
Where,
n = 1, 2, 3, …
According to de Broglie’s equation:
$\lambda=\frac{\text{h}}{\text{mv}}$
or $\text{mv}=\frac{\text{h}}{\lambda}...(2)$
Substituting the value of ‘mv’ from expression (2) in expression (1):
$\frac{\text{hr}}{\lambda}=\text{n}\frac{\text{h}}{2\pi}$
or $2\pi\text{r}=\text{n}\lambda...(3)$
Since $'2\pi\text{r}'$ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.
View full question & answer→Question 643 Marks
How many neutrons and protons are there in the following nuclei?
$\text{ }^{13}_{06}\text{C},\text{ }^{16}_{08}\text{O},\text{ }^{24}_{12}\text{Mg},\text{ }^{56}_{26}\text{Fe},\text{ }^{88}_{38}\text{Sr}$
Answer$\text{ }^{13}_{06}\text{C}:$
Atomic mass = 13
Atomic number = Number of protons = 6
Number of neutrons = (Atomic mass) – (Atomic number)
= 13 – 6 = 7
$\text{ }^{16}_{08}\text{O}:$
Atomic mass = 16
Atomic number = 8
Number of protons = 8
Number of neutrons = (Atomic mass) – (Atomic number)
= 16 – 8 = 8
$\text{ }^{24}_{12}\text{Mg}:$
Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) – (Atomic number)
= 24 – 12 = 12
$\text{ }^{56}_{26}\text{Fe}:$
Atomic mass = 56
Atomic number = Number of protons = 26
Number of neutrons = (Atomic mass) – (Atomic number)
= 56 – 26 = 30
$\text{ }^{88}_{38}\text{Sr}:$
Atomic mass = 88
Atomic number = Number of protons = 38
Number of neutrons = (Atomic mass) – (Atomic number)
= 88 – 38 = 50
View full question & answer→Question 653 Marks
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Answer Let the number of protons in the element be x.
$\therefore$ Number of neutrons in the element
= x + 31.7% of x
= x + 0.317 x
= 1.317 x
According to the question,
Mass number of the element = 81
$\therefore$ (Number of protons + number of neutrons) = 81
$\Rightarrow\text{x}+1.317\text{x}=81$
$2.317\text{x}=81$
$\text{x}=\frac{81}{2.317}$
$=34 .95$
$\therefore\text{x}=35$
Hence, the number of protons in the element i.e., x is 35.
Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.
$\therefore$ The atomic symbol of the element is $\text{ }^{81}_{35}\text{Br}.$
View full question & answer→Question 663 Marks
How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).
Answer $\text{E}_{\text{n}}=\frac{-21.8\times10^{-19}}{\text{n}^2\text{Jatom}^{-1}}$
For ionization from 5th orbit, $\text{n}_{1}=5,\text{n}_2=\infty$
$\therefore\triangle\text{E = E}_2-\text{E}_1=-21.8\times10^{-19}\times\Big(\frac{1}{\text{n}2^2}-\frac{1}{\text{n}1^2}\Big)$
$=21.8\times10^{-19}\times\Big(\frac{1}{\text{n}1^2}-\frac{1}{\text{n}2^2}\Big)$
$=21.8\times10^{-19}\times\Big(\frac{1}{5^2}-\frac{1}{\infty}\Big)$
$=8.72\times10^{-20}\text{J}$
For ionization from 1st orbit, $\text{n}_1=1,\text{n}_2=\infty$
$\therefore\triangle\text{E}'=21.8\times10^{-19}\times\Big(\frac{1}{1^2}-\frac{1}{\infty}\Big)=21.8\times10^{-19}\text{J}$
$\frac{\triangle\text{E}'}{\triangle\text{E}}=\frac{21.8\times10^{-19}}{8.72\times10^{-20}}=25$
Hence, 25 times less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.
View full question & answer→Question 673 Marks
Which of the following are isoelectronic species i.e., those having the same number of electrons?
$\text{Na}^+,\text{K}^+,\text{Mg}^{2+},\text{Ca}^{2+},\text{S}^{2-},\text{Ar}.$
AnswerNotes:Isoelectronic are the species having same number of electrons.
A positive charge means the shortage of an electron.
A negative charge means gain of electron.
Number of electrons in $\mathrm{Na}^{+}=11-1=10$
Number of electrons in $\mathrm{K}^{+}=19-1=18$
Number of electrons in $\mathrm{Mg}^{2+}=12-2=10$
Number of electrons in $\mathrm{Ca}^{2+}=20-2=18$
Number of electrons in $\mathrm{S}^{2-}=16+2=18$
Number of electrons in $\mathrm{Ar}=18$
Hence, the following are isoelectronic species:
1. $\mathrm{Na}^{+}$and $\mathrm{Mg}^{2+}$ (10 electrons each)
2. $\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-}$ and Ar (18 electrons each)
$\lambda=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)\left(811.579 \mathrm{~ms}^{-1}\right)}$
$\lambda=8.9625 \times 10^{-7} \mathrm{~m}$
Hence, the wavelength of the electron is $8.9625 \times 10^{-7} \mathrm{~m}$.
View full question & answer→Question 683 Marks
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition $n = 4 to n = 2 of He^+$ spectrum?
AnswerFor $He^+$ ion, the wave number $(\bar{\text{v}})$ associated with the Balmer transition, n = 4 to n = 2 is given by:$\bar{\text{v}}=\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Big)$
Where,
$n_1 = 2$
$n_2 = 4$
Z = atomic number of helium
$\bar{\text{v}}=\frac{1}{\lambda}=\text{R}(2)^2\Big(\frac{1}{4}-\frac{1}{16}\Big)$
$=4\text{R}\Big(\frac{4-1}{16}\Big)$
$\bar{\text{v}}=\frac{1}{\lambda}=\frac{3\text{R}}{4}$
$\Rightarrow\lambda=\frac{4}{3\text{R}}$
According to the question, the desired transition for hydrogen will have the same wavelength as that of $He^+$.
$\Rightarrow\text{R}(1)^2\bigg[\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}_2^2}\bigg]=\frac{3\text{R}}{4}$
$\bigg[\frac{1}{\text{n}_1^2}-\frac{1}{\text{n}^2_2}\bigg]=\frac{3}{4}...(1)$
By hit and trail method, the equality given by equation (1) is true only when
$n_1 = 1$ and $n_2 = 2.$
$\therefore$ The transition for $n_2 = 2 to n = 1$ in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of $He^+$ spectrum.
View full question & answer→Question 693 Marks
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is $2.5 \times 10^{15}$, calculate the energy of the source.
AnswerFrequency of radiation ( $v$ ),
$\mathrm{v}=\frac{1}{2.0 \times 10^{-9} \mathrm{~s}}$
$v=5.0 \times 10^8 \mathrm{~s}^{-1}$
Energy $(E)$ of source $=\mathrm{Nh} v$
Where,
$\mathrm{N}=$ number of photons emitted
$\mathrm{h}=$ Planck's constant
$v=$ frequency of radiation
Substituting the values in the given expression of (E):
$\mathrm{E}=\left(2.5 \times 10^{15}\right)\left(6.626 \times 10^{-34} \mathrm{~J}\right)\left(5.0 \times 10^8 \mathrm{~s}^{-1}\right)$
$E=8.282 \times 10^{-10} \mathrm{~J}$
Hence, the energy of the source $(\mathrm{E})$ is $8.282 \times 10^{-10} \mathrm{~J}$.
View full question & answer→Question 703 Marks
Threshold frequency, $v_0$ is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency $1.0 \times 10^{15} \mathrm{~s}^{-1}$ was allowed to hit a metal surface, an electron having $1.988 \times 10^{-19}$ J of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.
Answer$\text{hv}=\text{h}_0+\text{K.E}.$
$\text{hv}_0=\text{hv}-\text{K.E.}$
$\text{v}_0=\text{v}-\frac{\text{K.E.}}{\text{h}}\ ...(\text{i})$
$\text{v}=1.0\times10^{15}\text{s}^{-1}$
$\text{K.E.}=1.988\times^{-19}\text{J},\text{ h}=6.626\times10^{-34}\text{Js}$
From (i) we have,
$\text{v}_0=1.0\times10^{15}\text{s}^{-1}-\frac{1.988\times10^{-19}\text{J}}{6.626\times10^{-34}\text{Js}}$
$=(1.0\times10^{15}-0.30\times10^{15})\text{s}^{-1}$
$=0.7\times10^{15}\text{s}^{-1}=7\times10^{14}\text{s}^{-1}$
$\lambda=600\text{nm}=600\times10^{-9}\text{m}=6.0\times^{-7}\text{m}$
$\text{v}=\frac{\text{v}}{\lambda}=\frac{3.0\times10^8\text{ms}^{-1}}{6.0\times10^{-1}\text{m}}=5\times10^{14}\text{s}^{-1}$
Thus $\text{v}$ < $\text{v}_0$, Hence, no electron will be emitted.
View full question & answer→Question 713 Marks
Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it.
AnswerThe de-broglie relationship is given by,
$\lambda=\frac{\text{h}}{\text{mv}}$
Where $\lambda$ is wavelength, m is mass and v is velocity.
We can also write the relation as,
$\text{v}=\frac{\text{h}}{\text{m}\lambda}$
So, velocity is inversely proportional to the mass.
$\text { Mass of electron }=9.1 \times 10^{-31} \mathrm{~kg}$
$\text { mass of proton }=1.6 \times 10^{-27} \mathrm{~kg}$
Mass of proton is more than mass of electron therefore the velocity of electron is more than that of proton.
View full question & answer→Question 723 Marks
If the velocity of the electron in Bohr’s first orbit is $2.19 \times 106 \mathrm{~ms}^{-1}$, calculate the de Broglie wavelength associated with it.
AnswerAccording to de Broglie’s equation,
$\lambda=\frac{\text{h}}{\text{mv}}$
Where,
$\lambda$ = wavelength associated with the electron
h = Planck’s constant
m = mass of electron
v = velocity of electron
Substituting the values in the expression of $\lambda:$
$\lambda=\frac{6.626\times10^{-34}\text{Js}}{(9.10939\times10^{-31}\text{kg})(2.19\times10^6\text{ms}^{-1})}$
$=3.32\times10^{-10}\text{m}=3.32\times10^{-10}\text{m}\times\frac{100}{100}$
$=332\times10^{-12}\text{m}$
$\lambda=332\text{pm}$
$\therefore$ Wavelength associated with the electron = 332pm
View full question & answer→Question 733 Marks
What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
Answer
When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:
Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.
The number of spectral lines produced when an electron in the $\text{n}^{th}$ level drops down to the ground state is given by $\frac{\text{n(n}-1)}{2}$
Given,
n = 2
Number of spectral lines $=\frac{6(6-1)}{2}=15$ View full question & answer→Question 743 Marks
The electron energy in hydrogen atom is given by $\text{E}_{\text{n}}=\frac{ (–2.18 \times 10–18 )}{\text{n}^2}\text{J}.$ Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer $\text{E}_{\text{n}}=\frac{ (–2.18 \times 10–18 )}{\text{n}^2}\text{J}$
Energy required for ionization from n = 2 is given by,
$\triangle\text{E = E}_{\infty}-\text{E}_2$
$=\Big[\Big(\frac{-2.18\times10^{-18}}{(\infty)^2}\Big)-\Big(\frac{-2.18\times10^{-18}}{(2)^2}\Big)\Big]\text{J}$
$=\Big[\frac{2.18\times10^{-18}}{4}-0\Big]\text{J}$
$=0.545\times10^{-18}\text{J}$
$\triangle\text{E}=5.45\times10^{-19}\text{J}$
$\lambda=\frac{\text{hc}}{\triangle\text{E}}$
Here, $\lambda$ is the longest wavelength causing the transition.
$\lambda=\frac{(6.626\times10^{-34})(3\times10^8)}{5.45\times10^{-19}}=3.647\times10^{-7}\text{m}$
$=3647\times10^{-19}\text{m}$
$=3647\mathring{\text{A}}$
View full question & answer→Question 753 Marks
What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is $=–2.18\times10^{-11}\text{ ergs}$.
Answer1erg $=10^{-7} \mathrm{~J}$
As ground state electronic energy is $–2.18\times10^{-11}\text{ ergs}$, this means that $\text{E}_{\text{n}}\frac{-21.8\times10^{-11}}{\text{n}^2}\text{ergs}.$
$\triangle\text{E = E}_5-\text{E}_1=2.18\times10^{-11}\Big(\frac{1}{1^2}-\frac{1}{5^2}\Big)$
$=2.18\times10^{-11}\Big(\frac{24}{25}\Big)$
$=2.09\times10^{-1}\text{ ergs}$
$=2.09\times10^{-18}\text{J}$
When electron returns to ground state (n = 1), energy emitted $=2.09\times10^{-11}\text{ ergs}$.
As, $\text{E = hv}=\frac{\text{hc}}{\lambda}$
$\Rightarrow\lambda=\frac{\text{hc}}{\text{E}}=\frac{(6.626\times10^{-27}\text{ erg sec})(3.0\times10^{10}\text{cm s}^{-1})}{2.09\times10^{-11}\text{ergs}}$
$=9.51\times10^{-6}\text{cm}$
$=951\times10^{-8}\text{cm}$
$=951\mathring{\text{A}}$
View full question & answer→Question 763 Marks
Write the electronic configurations of the following ions :
- $H^–$
- $Na^+$
- $O^{2–}$
- $F^–$
Answer
- $H^– $
The electronic configuration of $H$ atom is $1s^1.$
A negative charge on the species indicates the gain of an electron by it.
$\therefore$ Electronic configuration of $H^– = 1s^2$
- $Na^+$
The electronic configuration of $Na$ atom is $1s^2 2s^2 2p^6 3s^1.$
A positive charge on the species indicates the loss of an electron by it.
$\therefore$ Electronic configuration of $Na^+ = 1s^2 2s^2 2p^6 3s^0$ or $1s^2 2s^2 2p^6$
- $O^{2–} $
The electronic configuration of $O$ atom is $1s^2 2s^2 2p^4.$
A dinegative charge on the species indicates that two electrons are gained by it.
$\therefore$ Electronic configuration of $O^{2–}$ ion $= 1s^2 2s^2 p^6$
- $F^–$
The electronic configuration of $F$ atom is $1s^2 2s^2 2p^5.$
A negative charge on the species indicates the gain of an electron by it.
$\therefore$ Electron configuration of $F^–$ ion $= 1s^2 2s^2 2p^6$ View full question & answer→Question 773 Marks
Yellow light emitted from a sodium lamp has a wavelength $(\lambda)$ of $580 nm$. Calculate the frequency (ν) and wavenumber $(\bar{\text{v}})$ of the yellow light.
AnswerFrom the expression,
$\lambda=\frac{\text{c}}{\text{v}}$
We get,
$\text{v}=\frac{\text{c}}{\lambda} \ ...(1)$
Where,
$v=$ frequency of yellow light
$\mathrm{c}=$ velocity of light in vacuum $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
$\lambda=$ wavelength of yellow light $=580 \mathrm{~nm}=580 \times 10^{-9} \mathrm{~m}$
Substituting the values in expression (i)
$\mathrm{v}=\frac{3 \times 10^8}{580 \times 10^{-9}}=5.17 \times 10^{14} \mathrm{~s}^{-1}$
Thus, frequency of yellow light emitted from the sodium lamp
$=5.17 \times 10^{14} \mathrm{~s}^{-1}$
Wave number of yellow light, $\bar{\text{v}}=\frac{1}{\lambda}$
$=\frac{1}{580\times10^{-9}}=1.72\times10^{6}\text{ m}^{-1}$
View full question & answer→