Question 12 Marks
What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?
AnswerEnergy $( E )=h v$
Energy of N photons $E = N h \nu= N h \frac{c}{\lambda}$
Hence $N =\frac{E \lambda}{h c}$
$E =1 J, \quad \lambda=4000 Pm =4000 \times 10^{-12} m$
$h=6.626 \times 10^{-34} J$ and $c=3 \times 10^8 ms^{-1}$
No. of photons (N)
$=\frac{1 J \times 4000 \times 10^{-12} m}{6.626 \times 10^{-34} Js \times 3 \times 10^8 ms^{-1}}$
$=2.012 \times 10^{16}$ Photon
View full question & answer→Question 22 Marks
Calculate the wavelength, frequency and wavenumber of a light wave whose period is $2.0 \times$ $10^{-10} s$.
AnswerTime period $(T)=2 \times 10^{-10} s$
(i) Frequency $v=\frac{1}{\text { Time period (T) }}$
$=\frac{1}{2 \times 10^{-10} s}=5 \times 10^9 s^{-1}$
(ii) Wavelength $\lambda=\frac{c}{v}=\frac{3 \times 10^8 ms^{-1}}{5 \times 10^9 s^{-1}}$
$\lambda=6.0 \times 10^{-2} m$
(iii) Wave no. $\bar{v}=\frac{1}{\lambda}=\frac{1}{6.0 \times 10^{-2} m}$
$=16.66 m^{-1}$
View full question & answer→Question 32 Marks
(a) How many subshells are associated with $n =4$ ? (b) How many electrons will be present in the subshells having $m_s$ value of $-\frac{1}{2}$ for $n=4$ ?
Answer(a) For $n=4, l=0,1,2,3$
\begin{array}{rlc}
& & \text { subshell } \\
l=0 & = & 4 s \\
1 & = & 4 p \\
2 & = & 4 d \\
3 & = & 4 f
\end{array}
Hence for $n=4$ there will be four subshells ( $4 s$, $4 p, 4 d$ and $4 f)$
(b) Number of orbitals for $n=4=4^2=16$ and for one electron present in each orbital $m _{ s }=-\frac{1}{2}$. Therefore the subshells with $n =4, m_{ s }=-\frac{1}{2}$ have 16 electrons.
View full question & answer→Question 42 Marks
Find energy of each of the photons which
(i) correspond to light of frequency $3 \times 10^{15} Hz$
(ii) have wavelength of $0.50 Å$
Answer(i) Energy of Photon (E) $=h v$
$h=$ Planck's constant $=6.626 \times 10^{-34} Js$
Hence, $E =6.626 \times 10^{-34} Js \times 3 \times 10^{15} s^{-1}$
$E=1.9878 \times 10^{18} J$
$E =1.988 \times 10^{-18} J$
(ii) Energy (E) $=h v=\frac{h c}{\lambda}$
$c=3 \times 10^8 ms^{-1}, \lambda=0.5$Å
$=0.5 \times 10^{-10} m$
$E =\frac{6.626 \times 10^{-34} Js \times 3 \times 10^8 ms^{-1}}{0.5 \times 10^{-10} m}$
$=3.9756 \times 10^{-15} J$
Energy $=3.98 \times 10^{-15} J$
View full question & answer→Question 52 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 800 pm , calculate the characteristic velocity associated with the neutron.
AnswerWavelength $(\lambda)=800 Pm =800 \times 10^{-12} m$
$=8 \times 10^{-10} m$
Mass of neutron $=1.675 \times 10^{-27} kg$
$\lambda=\frac{h}{m v } \quad$ or $\quad v =\frac{h}{m \lambda}$
$=\frac{6.626 \times 10^{-34} kg m ^2 s^{-1}}{1.675 \times 10^{-27} kg \times 8 \times 10^{-10} m}$
$\begin{array}{l}\lambda=494.47 ms^{-1} \\ \lambda=494.5 ms^{-1}\end{array}$
View full question & answer→Question 62 Marks
Emission transitions in the Paschen series end at orbit $n=3$ and start from orbit $n$ and can be represented as $v=3.29 \times 10^{15}(Hz)\left[\frac{1}{3^2}-\frac{1}{n^2}\right]$
Calculate the value of $n$ if the transition is observed at 1285 nm . Find the region of the spectrum.
AnswerFrequency $(v)=\frac{c}{\lambda},(\lambda=1285 nm$$\left.=1285 \times 10^{-9} m\right)$
$3.29 \times 10^{15}\left(\frac{1}{3^2}-\frac{1}{n^2}\right)=\frac{3 \times 10^8 ms^{-1}}{1285 \times 10^{-9} m}$
$\frac{1}{n^2}=\frac{1}{9}-\frac{3 \times 10^8}{1285 \times 10^{-9}} \times \frac{1}{3.29 \times 10^{15}}$
$\frac{1}{n^2}=0.111-0.071$
$\frac{1}{n^2}=0.04=\frac{1}{25}$
$n^2=25, \quad n=5$
This radiation is in infra red region.
View full question & answer→Question 72 Marks
If the photon of the wavelength 150 pm strikes an atom and one of this inner bound electrons is ejected out with a velocity of $1.15 \times 10^7 m s ^{-1}$, calculate the energy with which it is bound to the nucleus.
AnswerWavelength $\lambda=150 Pm =150 \times 10^{-12} m$
Energy of photon $( E )=h v=\frac{h c}{\lambda}$
$\begin{array}{l}=\frac{6.626 \times 10^{-34} Js \times 3 \times 10^8 ms^{-1}}{150 \times 10^{-12} m} \\ =1.32 \times 10^{-15} J\end{array}$
K.E. of emitted electron $=\frac{1}{2} m v^2$
$=\frac{1}{2} \times 9.11 \times 10^{-31} \times\left(1.5 \times 10^7\right)^2$
$=1.025 \times 10^{-16} J=0.1025 \times 10^{-15} J$
Binding energy $=$ Energy of incident photon -
Kinetic energy (K.E.)
$=1.32 \times 10^{-15} J-0.1025 \times 10^{-15} J$
Binding energy $=1.217 \times 10^{-15} J$
$=\frac{1.217 \times 10^{-15}}{1.6 \times 10^{-19}}=7.6 \times 10^3 eV$
View full question & answer→Question 82 Marks
The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm . Calculate the frequency of each transition and energy difference between two excited states.
Answer$\begin{array}{l}\lambda_1=589 nm=589 \times 10^{-9} m \\ \lambda_2=589.6 nm=589.6 \times 10^{-9} m\end{array}$
$v_1=\frac{ c }{\lambda_1}=\frac{3 \times 10^8 ms^{-1}}{589 \times 10^{-9} m}$
$v_2=\frac{ c }{\lambda_2}=\frac{3 \times 10^8 ms^{-1}}{589.6 \times 10^{-9} m}$
$\begin{array}{l}v_1=5.0933 \times 10^{14} s^{-1} \\ v_2=5.0881 \times 10^{14} s^{-1}\end{array}$
Energy difference
$
\begin{aligned}
\Delta E & =E_2-E_1 \\
\Delta E & =h v_2-h v_1 \\
\Delta E & =h\left(v_2-v_1\right) \\
& =6.626 \times 10^{-34} Js(5.0933-5.0881) \\
& \times 10^{14} s^{-1}
\end{aligned}
$
$\begin{array}{l}=6.626 \times 10^{-34} \times 0.0523 \times 10^{14} \\ =3.46 \times 10^{-22} J\end{array}$
View full question & answer→Question 92 Marks
In astronomical observation, signals observed from the distant stars are generally weak. If the photon detector receives a total of $3.15 \times 10^{-18}$ $J$ from the radiations of 600 nm , calculate the number of photons received by the detector.
AnswerWavelength $\lambda=600 nm=600 \times 10^{-9} m$, Energy of photon $=3.15 \times 10^{-18} J$
Energy of 1 photon, $E =h v=\frac{h c}{\lambda}$
Energy of $n$ photon $E =\frac{n h c}{\lambda}$
No. of photon, $n=\frac{E \lambda}{h c}$
$=\frac{3.15 \times 10^{-18} J \times 600 \times 10^{-9} m}{6.626 \times 10^{-34} Js \times 3 \times 10^8 ms^{-1}}$
Hence, No. of photons $=9.5=10$
View full question & answer→Question 102 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.
AnswerOn ion there is three unit positive charge. Hence in this number of electrons is three less than protons.
Hence mass number $(56)=$ Number of protons + Number of neutrons
$
(56-3)=e+n
$
Let number of electrons in ion $=x$
Hence neutron $=\frac{x+30.4 x}{100}=1.304 x$
But in ions sum of number of electrons and neutrons is $=53$
Hence $x+1.304 x=53$
$2.304 x=53$
$x=\frac{53}{2.304}$
$x=23=$ No. of electrons
Proton $=23+3=26$. Hence this element is ironhaving symbol ${ }_{26}^{56} Fe ^{3+}$.
View full question & answer→Question 112 Marks
An element with mass number 81 contains $31.7 \%$ more neutrons as compared to protons. Assign the atomic symbol.
AnswerMass no. $(81)= P + N$
Let no. of protons $=x$
According to question, no. of n
$
=\frac{x+31.7 x}{100}=1.317 x
$
Mass no. (81) $=x+1.317 x$
$2.317 x=81$
$\begin{array}{l}x=\frac{81}{2.317} \\ x=34.958=35\end{array}$
No. of protons $=35$, therefore element will be bromine and atoms symbol $={ }_{35}^{81} Br$
View full question & answer→Question 122 Marks
The diameter of zinc atom is $2.6 Å$. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
Answer(a) Diameter of Zn atom =2.6 Å
Hence radius $=\frac{2.6}{2}=1.3 Å$
$\begin{array}{l}=1.3 \times 10^{-10} m \\ =1.3 \times 10^{-10} \times 10^{12} Pm \\ =130 Pm \\ =1.3 \times 10^2 Pm \end{array}$
(b) Length $(l)=1.6 cm=1.6 \times 10^{-2} m$
For Zn diameter of 1 atom (d)=2.6 Å
$=2.6 \times 10^{-10} m$
Hence no. of atoms in $Zn n=\frac{l}{d}$
$\begin{array}{l}=\frac{1.6 \times 10^{-2} m}{2.6 \times 10^{-10} m} \\ =6.15 \times 10^7\end{array}$
View full question & answer→Question 132 Marks
$2 \times 10^8$ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm .
AnswerNo. of carbon atoms $(n)=2 \times 10^8$
Length of arrangement $(l)=2.4 cm$
$\begin{array}{l}=2.4 \times 10^{-2} m \\ =2.4 \times 10^7 nm\end{array}$
Diameter of carbon atom $(d)=\frac{l}{n}$
$\begin{array}{l}=\frac{2.4 \times 10^7 nm}{2 \times 10^8} \\ =0.12 nm\end{array}$
Radius of C atom $r=\frac{d}{2}=\frac{0.12}{2}$
$=0.06 nm$
View full question & answer→Question 142 Marks
Calculate the energy required for the process
$
He^{+}(g) \rightarrow He^{2+}(g)+e^{-}
$
The ionization energy for the $H$ atom in the ground state is $2.18 \times 10^{-18} J atom ^{-1}$.
AnswerFor $H _2$ atom
$E _1=- I . E .=-2.18 \times 10^{-18} J$
For $He ^{+} E _1= E _{1( H )} \times Z^2$
$=-2.18 \times 10^{-18} \times 2^2 J$
$=-8.72 \times 10^{-18} J$
In $He _{(g)}^{+}\left(1 s^1\right) \longrightarrow He ^{2+}+e$ ionisation of $He ^{+}$is happening.
$\begin{aligned} \text { Hence } & \text { energy required }=- E _{1\left( Hc ^{+}\right)} \\ & =-\left(-8.72 \times 10^{-18}\right) J \\ & =8.72 \times 10^{-18} J \text { per atom }\end{aligned}$
View full question & answer→Question 152 Marks
How many electron in an atom may have the following quantum numbers?
(a) $n =4, m_s=-1 / 2$
(b) $n =3, l=0$.
Answer(a) $n =4$ i.e. in fourth shell total numbers of electrons $=2 n ^2=2 \times 4^2=32$.
Out of these electrons for 16 electrons $m_s=-1 / 2$ and for remaining $16 m_{ s }=+1 / 2$
Hence there are 16 electrons having $n =4$ and $m _s$ $=-1 / 2$
(b) For $n =3$ and $l=0$, it will be 3 s orbital in which two electrons can accomodate because the maximum electron capacity of any orbital $=2$.
View full question & answer→Question 162 Marks
How many neutrons and protons are there in the following nuclei?
${ }_6^{13} C ,{ }_8^{16} O ,{ }_{12}^{24} Mg ,{ }_{26}^{56} Fe ,{ }_{38}^{88} Sr$
AnswerNumber of neutrons $( n )=$ Mass number $( A )$- Atomic number (z)
Number of protons $=$ Atomic number $( z )$| Nucleus | Mass No. (A) | Atomic no. (No. of Protons) (Z) | No. of neutrons$=$ Mass No. -- Atomic No.$(n= A - Z )$ |
| ${ }_6^{13} C$ | 13 | 6 | $(13-6)=7$ |
| ${ }_8^{16} O$ | 16 | 8 | $(16-8)=8$ |
| ${ }_{12}^{24} Mg$ | 24 | 12 | $(24-12)=12$ |
| ${ }_{26}^{56} Fe$ | 56 | 26 | $(56-26)=30$ |
| ${ }_{38}^{88} Sr$ | 88 | 38 | $(88-38)=50$ |
View full question & answer→Question 172 Marks
Which of the following are isoelectronic species i.e., those having the same number of electrons?
$
Na^{+}, K^{+}, Mg^{2+}, Ca^{2+}, S^{2-}, Ar
$
Answer(i) Electrons in $Na ^{+}=11-1=10$
(ii) $K ^{+}=19-1=18$ electrons
(iii) $Mg ^{2+}=12-2=10$ electrons
(iv) $Ca ^{2+}=20-2=18$ electrons
(v) $Ar =18$ electrons
(vi) $S ^{-2}=16+2=18$ electrons
Hence $Na ^{+}$and $Mg ^{+2}$ are isoelectronic and $K ^{+}, Ca ^{2+}$, Ar and $S ^{2-}$ are isoelectronic.
View full question & answer→Question 182 Marks
The mass of an electron is $9.1 \times 10^{-31}$ kg. If its $K . E$ is $3.0 \times 10^{-25} J$, calculate its wavelength.
AnswerKinetic Energy (K.E.) $=\frac{1}{2} m v^2$
K.E. $=3 \times 10^{-25} J$, Mass of electron
$
m=9.1 \times 10^{-31} kg
$
Hence, velocity of electron $(v)=\left[\frac{2 \times \text { K.E. }}{m}\right]^{\frac{1}{2}}$
$v =\left(\frac{2 \times 3.0 \times 10^{-25}}{9.1 \times 10^{-31}}\right)^{\frac{1}{2}}=0.8119 \times 10^3 ms^{-1}$
$=0.812 \times 10^3 ms^{-1}$
Wavelength,
$\begin{array}{l}\lambda=\frac{h}{m V}=\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 0.812 \times 10^3} \\ \lambda=0.8967 \times 10^{-6} m \\ \lambda=0.8967 \times 10^{-6} \times 10^{10} Å \\ \lambda=8967 Å\end{array}$
View full question & answer→Question 192 Marks
The electron energy in hydrogen atom is given by $E_n=\left(-2.18 \times 10^{-18}\right) / n ^2 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?
View full question & answer→Question 202 Marks
Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
AnswerWavenumber, $\bar{v}=\frac{1}{\lambda}$, hence when wavelength is maximum, wavenumber $\bar{v}$ and energy will be minimum because :
$
E=h v
$
For Balmer series $n_1=2, n_2=3$ (Minimum $\bar{v}$ )
$\bar{v}= R \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) cm ^{-1}$
$\bar{v}=109677\left(\frac{1}{2^2}-\frac{1}{3^2}\right) cm ^{-1}$
$\bar{v}=109677\left(\frac{1}{4}-\frac{1}{9}\right) cm ^{-1}$
$\bar{v}=109677\left(\frac{5}{36}\right) cm ^{-1}$
$\bar{v}=1.52329 \times 10^4 cm^{-1}$
$\bar{v}=1.5233 \times 10^6 m^{-1}$
View full question & answer→Question 212 Marks
(i) The energy associated with the first orbit in the hydrogen atom is $-2.18 \times 10^{-18} J$ atom ${ }^{-1}$. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr's fifth orbit for hydrogen atom.
Answer(i) Energy of $n ^{\text {th }}$ orbital in hydrogen
$=\frac{-2.18 \times 10^{-18}}{n^2} J$
$n=5$
Hence, $E_5=\frac{-2.18 \times 10^{-18}}{5^2}=8.72 \times 10^{-20} J$
(ii) Radius of $n ^{\text {th }}$ Bohr orbital of hydrogen atom
$
r_n=0.529 \times n^2 Å
$
Hence, $\quad \begin{aligned} n & =5 \\ r_5 & =0.529 \times 5^2=13.225 Å \\ & =1.3225 nm\end{aligned}$
View full question & answer→Question 222 Marks
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with $n=4$ to an energy level with $n =2$ ?
AnswerWave number of hydrogen $(\bar{v})$
$= R \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \quad n_2>n_1$
Here, $n _2> n _1$
According to question
$R =109677 cm^{-1}, \quad n_2=4, \quad n_1=2$
$\bar{v}=109677\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$
$\bar{v}=109677 \times \frac{3}{16}$
$\bar{v}=20564.4 cm^{-1}$
Wavelength $\lambda=\frac{1}{\bar{v}}=\frac{1}{20564.4} cm$
$\lambda=486 \times 10^{-7} cm$
$\lambda=486 \times 10^{-9} m$
$\lambda=486 nm$
View full question & answer→Question 232 Marks
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength $6800 Å$. Calculate threshold frequency $\left(v_0\right)$ and work function $\left( W _0\right)$ of the metal.
AnswerWavelength $\lambda_0=6800 Å=6800 \times 10^{-10} m$
Threshold frequency $v_0=\frac{c}{\lambda_0}=\frac{3 \times 10^8 ms^{-1}}{6800 \times 10^{-10} m}$
$=4.41 \times 10^{14} s^{-1}$
Work function $=h v_0$
Because when velocity is zero, kinetic energy is also zero.
Therefore wave function
$W _0=6.626 \times 10^{-34} \times 4.41 \times 10^{14}$
$=2.92 \times 10^{-19} J$
View full question & answer→Question 242 Marks
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 µm. Calculate the rate of emission of quanta per second.
AnswerWavelength $(\lambda)=0.57 \quad \mu m=0.57 \times 10^{-6} m$
Power of bulb $=25$ Watt $=25 Js ^{-1}$
Energy of photon $( E )=h v=\frac{h c}{\lambda}$
$=\frac{6.626 \times 10^{-34} Js \times 3 \times 10^8 ms^{-1}}{0.57 \times 10^{-6} m}$
$=3.48 \times 10^{-19} J$
Rate of emmision of quanta per second
$
\begin{array}{l}
=\frac{\text { Power of bulb }}{\text { Energy of photon }} \\
=\frac{25 Js^{-1}}{3.48 \times 10^{-19}} \\
=7.18 \times 10^{19} s^{-1}
\end{array}
$
View full question & answer→Question 252 Marks
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ $mol ^{-1}$.
AnswerFor ionization of sodium atom, the electromagnetic radiation of 242 nm wavelength is sufficient. Hence ionization energy = energy of radiation
$\lambda=242 nm=242 \times 10^{-9} m$
Ionisation energy $( E )=h v$
$=\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} Js \times 3 \times 10^8 ms^{-1}}{242 \times 10^{-9} m}$
$=8.21 \times 10^{-19} J /$ photon
$=8.21 \times 10^{-22} kJ /$ photon
I.E. of 1 mole $Na =$ I.E. $\times N _{ A }$
$=8.21 \times 10^{-22} kJ \times 6.022 \times 10^{23}$
$=494 kJ mol ^{-1}$
View full question & answer→Question 262 Marks
Explain the characteristics of canal rays (positively charged particles).
AnswerFollowing are the characteristics of canal rays :
(i) Positively charged particles depend on the nature of the gas present in the cathode ray tube and they are simple positively charged gaseous ions.
(ii) The ratio of charge and mass of the particles depends on gas from which they are produced.
(iii) The electric charge of some positively charged particles is a simple multiple of fundamental unit.
(iv) The behavior of these particles in magnetic and electric fields is opposite to the behaviour of electron or cathode ray.
(v) The mass of each particle of positive rays is approximately equal to the atomic weight of the gas used in the discharge tube.
View full question & answer→Question 272 Marks
Find the frequency of energy required when the electron moves from the first orbit to be the third orbit in the hydrogen atom.
AnswerFrequency
$
\begin{array}{c}
(V)=3.29 \times 10^{15}\left[\frac{1}{n_{i}^2}-\frac{1}{n_{f}^2}\right] Hz \\
V=3.29 \times 10^{15}\left[\frac{1}{1^2}-\frac{1}{3^2}\right] \\
V=3.29 \times 10^{15} \times \frac{8}{9} \\
V=2.9244 \times 10^{15} Hz
\end{array}
$
View full question & answer→Question 282 Marks
When the electron in hydrogen atom $n =1$ if it travels in $n =2$ then what will be the required energy ( eV )?
Answer$\quad$ Energy $E=-13.6 \frac{z^2}{n^2} eV$.
$
Z=1
$
Energy of first orbital $\left(E_1\right)=\frac{-13.6 \times 1^2}{1^2}=-13.6 eV$
Energy of Second orbital ( $E _2$ )$
\begin{aligned}
& =\frac{-13.6 \times 1^2}{2^2}=-3.4 eV \\
\text { Hence } E_2-E_1 & =-3.4-(-13.6) eV \\
& =10.2 eV
\end{aligned}
$
View full question & answer→Question 292 Marks
The frequency of any radiation is $4 \times 10^{10}$ $s ^{-1}$ then calculate the energy of one mole photons.
AnswerEnergy of one photon $(E)= h V$$
=6.62 \times 10^{-34} Js \times 4 \times 10^{10} s^{-1}
$
$
=26.48 \times 10^{-24} J
$
Energy of one mole photons
$=$ Energy of one photon $\times N _{ A }$
$=26.48 \times 10^{-24} J \times 6.022 \times 10^{23}$
$=15.94 J$
View full question & answer→Question 302 Marks
If the uncertainty in the velocity of an electron is $6 \times 10^5 ms^{-1}$ then find the uncertainty in its position.
Answer
$
\Delta x \times \Delta p=\frac{h}{4 \pi}
$
Mass of electron $(m)=9.1 \times 10^{-31} kg$$
h=6.62 \times 10^{-34} kg m^2 s^{-1}
$
and $\Delta V =6 \times 10^5 ms^{-1}$
Hence uncertainty in position of electrons$
\begin{array}{c}
\Delta x=\frac{h}{4 \pi \Delta p}=\frac{h}{4 \pi m \Delta V} \\
=\frac{6.62 \times 10^{-34} kg m^2 s^{-1}}{4 \times 3.14 \times 9.1 \times 10^{-31} kg \times 6 \times 10^5 ms^{-1}} \\
\Delta x=9.6 \times 10^{-11} m
\end{array}
$
View full question & answer→Question 312 Marks
Calculate the wavelength of one photon whose energy is 1 eV .
AnswerEnergy $(E)=h V=\frac{h c}{\lambda}$
Energy, $E =1 eV =1.6 \times 10^{-19} J$
Hence
$\begin{array}{l}
\lambda=\frac{hc}{E}=\frac{6.62 \times 10^{-34} Js \times 3 \times 10^8 ms^{-1}}{1.6 \times 10^{-19} J} \\
\lambda=12.41 \times 10^{-7} m \\
\lambda=12.41 \times 10^{-7} \times 10^{10} Å \\
\lambda=12.41 \times 10^3 Å
\end{array}
$
View full question & answer→Question 322 Marks
If the value of energy related to the first orbital of hydrogen is $-13.6 eV /$ atom, then find the energy related to the third orbital.
AnswerEnergy of any orbital in atom $\left(E_n\right)$
$
E_{n}=\frac{E_1}{n^2}
$
$E _1=$ Energy of first orbital and $n =$
value of orbital
$
E_3=\frac{-13.6}{3^2}=-1.51 eV
$
View full question & answer→Question 332 Marks
Define the following and also tell their units.
(i) Frequency
(ii) Wavelength
(iii) Wavenumber.
Answer(i) Frequency ( $V$ ) : The number of waves passing through a point per second is called frequency. Its unit is Hertz (Hz) or second ${ }^{-1}\left(s^{-1}\right)$
(ii) Wavelength ( $\lambda$ ) : The distance between the two nearest crests or troughs of a wave is called wavelength. Its unit is meter or cm or Å.
(iii) Wavenumber ( $\bar{V}$ ) : The number of wavelengths per unit length is called wavenumber.
$
\bar{V}=\frac{1}{\lambda}
$
unit of $\bar{V}= m ^{-1}$ or $cm ^{-1}$
View full question & answer→Question 342 Marks
(a) If there is only one electron in the 3 p subshell, then in which of the $3 p _x, 3 p _y$ and $3 p _z$ orbitals will this electron be present?
(b) What are the quantum numbers representing the following :
(i) Size of orbitals
(ii) Shape of orbitals
(iii) Orientation of orbitals
(c) If the mass number of an element is double its atomic number and it has four electrons in its 2 p subshell, then write the name and electronic of this element.
Answer(a) An electron present in the 3 p subshell can be present in any of the $3 p _x, 3 p _y$ and $3 p _z$ orbitals because the energy of all these orbitals is the same.
(b) (i) Principal quantum number
(ii) Azimuthal quantum number
(iii) Magnetic quantum number
(c) This element is oxygen whose mass number (16) which is doubled that atomic number (8). Its electronic configuration is as follows : ${ }_8 O =1 s^2, 2 s^2$, $2 p^4$ whose $2 p$ subshell have four electrons.
View full question & answer→Question 352 Marks
(a) Name one element in which the number of $s$ electrons is equal to the numbers of p-electrons.
(b) Tell the number of radial nodes in 1 s and 2 s orbitals.
Answer(a) Number of s and p electrons in magnesium is equal :
$
\begin{array}{l}
{ }_{12} Mg=1 s^2, 2 s^2, 2 p^6, 3 s^2(s \text { electron }= 6 \\
p \text { electron }=6)
\end{array}
$
(b) Number of radial nodes $= n -1$
Hence 1 s and 2 s , the number of radial nodes in the orbitals is zero and one.
View full question & answer→Question 362 Marks
(a) If a moving proton and an electron have the same wavelength, then whose velocity will be greater and why?
Answer(a) According to formula $\lambda=\frac{ h }{ mV }$, when $\lambda$ and h are constant then velocity is inversely proportional to mass, as mass of electron is less than that of proton, hence its velocity will be more.
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Explain the statement that linear spectrum of an element is called fingerprint of that element.
Answer Each element has its own special linear spectrum, hence it is used to identify unknown atoms in the same way as a person is identified through his fingerprint, hence the linear spectrum is called the fingerprint of that element.
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Explain the Heisenberg's uncertainity principle.
Question 392 Marks
(a) A rotating cricket ball does not have wave nature, why?
(b) Why should one not wear black clothes in summer?
Answer(a) According to formula $\lambda=\frac{ h }{ mV }$, the object having more mass, like cricket ball etc., is so short that it is not possible to measure it easily. Therefore, a rotating cricket ball does not have wave nature.
(b) Black clothes behave like a black body hence they generate heat by absorbing the radiations received from the sun, hence black clothes should not be worn in summer.
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Explain Schrodinger wave equation.
Question 412 Marks
Difterentiate between electromagnetic wave and matter wave.
Question 422 Marks
Write the electronic configuration of following elements :
(i) ${ }_{29} Cu$(ii) ${ }_{42} Mo$(iii) ${ }_{46} Pd$
(iv) ${ }_{47} Ag$(v) ${ }_{78} Pt$(vi) ${ }_{79} Au$
Answer(i) ${ }_{29} Cu =[ Ar ] 3 d^{10} 4 s^1$
(ii) ${ }_{42} Mo =[ Kr ] 4 d^5 5 s^1$
(iii) ${ }_{46} Pd =[ Kr ] 4 d^{10} 5 s^0$
(iv) ${ }_{47} Ag =[ Kr ] 4 d^{10} 5 s^1$
(v) ${ }_{78} Pt =[ Xe ] 4 f^{14} 5 d^9 6 s^1$
(vi) ${ }_{79} Au =[ Xe ] 4 f^{14} 5 d^{10} 6 s^1$
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What are nodal planes and what will be the number of nodal planes for 4 s orbital?
Answer The region where the electron probability density is zero i.e. the space around the nucleus where the probability of finding an electron is almost zero, is called nodal plane or radius node. Nodal surface for S orbital $= n =1$. So there will be 3 nodal planes for 4 s .
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What is the maximum number of electrons in the following :
(i) Shell (ii) M shell (iii) Subshell (iv) d subshell
Answer (i) Maximum number of electrons in any shell $=2 n ^2$
(ii) For M shell, $n =3$
Therefore, maximum number of electron in it $=$ $2 n ^2=2(3)^2=18$
(iii) The maximum number of electrons in any subshell $=2(2 l+1)$
(iv) For d subshell $l=2$ therefore in this maximum number of electrons
$
\begin{array}{l}
=2(2 l+1) \\
=2(2 \times 2+1)=10
\end{array}
$
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There is only one electron in a hydrogen atom, yet many lines are visible in its spectrum, why?
Answer An atom of hydrogen can be in only one excited state at any one time, but while taking the spectrum, there is a group of atoms which has all the possible excited states, hence when these atoms come to the lower energy state, then is emitted spectrum many lines are visible simultaneously.
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Why is the atomic spectrum is not continuous but linear?
AnswerWhen continuous radiation is passed through a substance, it absorbs some wavelengths of radiation. The missing wavelengths corresponding to the radiation absorbed by this substance appear as dark lines in the bright continuous spectrum, because only light of particular wavelengths is emitted with dark spaces in between. Hence such spectrum is called linear spectrum. It is displayed by atoms only in gaseous state, hence it is also called atomic spectrum.
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What is photoelectric effect and also tell the results of its experiment?
Answer When light of appropriate frequency attack on reactive metals like $K , Rb$ and Cs etc., the ejection of electrons from their surface is called photoelectric effect and the emitted electrons are called photoelectrons.
[Hint : See the Point no. 2.3.2 of the Text.]
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Write the formula to calculate the radius of Bohr orbital of hydrogen atom.
Answer$\begin{aligned} r_n & =\frac{n^2 h^2}{4 \pi^2 m e^2} \\ r_n & =n^2 \times 0.529 \times 10^{-10} m\end{aligned}$
$r_n=n^2 \times 0.529$Å
$r_n=n^2 a_0$
$a_o=$ Bohr radius $=0.529$Å
$a_o=52.9 P_m$
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Explain Thomson Model of atom.
AnswerAccording to J.J. Thomson (1898), atom is a charged sphere, on which the positive charge is distributed thoroughly and the electrons are embedded in it in way that a stable constant electric arrangement is obtained. This is known as Plum pudding model or watermelon model. In this positive charge is considered as pudding or watermelon and electron is considered as plum or seed.
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How was the neutron discovered and what are its properties?
AnswerChadwick (1932) discovered neutrons by the bombardment of alpha ( $\alpha$ ) particles on Be . It is a neutral particle which is found in the nucleus of the atom and its mass is slightly more than the proton.
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Explain magnetic quantum number.
AnswerMagnetic quantum number $\left(m_1\right)$ :
(i) It describes about orbital.
(ii) It gives information about 3D configuration of orbitals.
(iii) Number of orbitals in subshell $=2 l+1=$ possible values of $m_l$.
(iv) $m_l=-l,-(l-1),-(l-2), \ldots \ldots \ldots 0$,$\ldots \ldots \ldots (l-2),(l-1), l$
(v) $l=0(s) m_l=0$, one $s$-orbital
$\begin{aligned} l & =1(p) m_l=-1,0,+1, \text { three } p \text {-orbital } \\
l & =2(d) m_l=-2,-1,0,+1,+2, \text { five } d \text {-orbital } \\
l & =3(f) m_l=-3,-2,-1,0,+1,+2,+3, \text { seven }\end{aligned}$
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(a) What will be the effect on the wavelength of a moving particle if its velocity is doubled?
(b) The spin of two electrons present in an orbital is opposite, why?
Answer(a) On doubling the velocity the wavelength will be halved because $\lambda=\frac{ h }{ mV }$ and $\frac{ h }{ m }$ is constant.
(b) When two electrons present in our orbital have opposite spins, their repulsion is less.
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