5 Marks Questions

Question 15 Marks

Explain the shortcomings of Rutherford's Atomic Model.###Write two shortcomings of Rutherford's Atomic Model.

Answer

Drawbacks of Rutherford's Nuclear Atomic Model : This model suffers from two drawbacks :
(i) Regarding stability of atom : There is centripetal acceleration in the revolving electron around the nucleus. According to electromagnetic theory, accelerated charged particle emits energy. Hence, electromagnetic waves should continuously emit by the revolving electrons in different orbits around the nucleus. In this way due to loss of energy of electron, radius of circular orbit should continuously decreases and ultimately the electrons should fall into the nucleus. Therefore atom cannot remain stable.

Fig. : Rutherford's Atomic Model pattern
(ii) Regarding explanation of Line-spectrum : In Rutherford's model, the frequency of moving electrons will changes due to continuously changing in radius of electrron's path. After this result, electrons will emit electromagnetic waves of all frequencies i.e. the spectrum of the waves will be continuous, but in really the atomic spectra have line spectrum instead of continuous and each spectral line have definite frequency. Therefore emitted wave from atom should be of some definite frequencies, instead of all frequencies. Thus, Rutherford model unable explain the line spectrum.

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Question 25 Marks

Write first and second postulates of Bohr's Atomic Model. Obtain the expression for radius and velocity of stable orbit of electron.###Explain Bohr's two postulates for hydrogen atom.

Answer

Bohr's postulates : Bohr proposed three postulates which are following :
(1) Bohr's first postulates : Any electron can revolve in the definite stable states of atom without radiating the emission energy. It is opposite to the electromagnetic principle. According to this postulate, each atom has same definite stable states in which it existed and existed total energy is definite in all possible states. All these possible states are called stable states of atom.
(2) Bohr's second postulate : Second postulate defines to these stable orbit. According to this postulate, electron revolves around the nucleus only in those orbits for which angular momentum is multiple integer of $\frac{h}{2 \pi}$. Where, $h$ Planck's constant $\left(=6.6 \times 10^{-34} Js \right)$. Therefore, L is quantised angular momentum of revolving electron. That is,
$L=\frac{n h}{2 \pi}$ $\ldots (1)$
(3) Bohr's third postulate : According to this postulate, transition of electron can be done from specially given unradiated orbits to second low energy orbits. When it does so a photon is emitted, whose energy is equal to the difference of energy of initial and final states. Frequency of emitted photon is given by following expression :
$h v=E_i-E_f$ $\ldots (2)$
Where, $E _i$ and $E _f$ are the energies of initial and final states, $E _i> E _f$
Bohr's Radius or radius of stable orbits : Let $r_n$ is the radius of $n^{\text {th }}$ orbit of any atom and in which an electron is moving with velocity $v_n$. If mass of electron is $m$ then from the Bohr's second postulate,
$m v_n r_n=n \cdot \frac{h}{2 \pi}$ $\ldots (1)$

From Bohr's first postulate,
$F _e = F _c$
$\therefore \frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n^2}=\frac{m v^2 n}{r_n}$ $\ldots (2)$
From eqn. (1) $v_n=\frac{n h}{2 \pi m r_n}$
Put value of $v_n$ in eqn. (2)
$\frac{1}{4 \pi \epsilon_0} \cdot \frac{ Z e^2}{r_n^2}=\frac{m}{r_n} \times \frac{n^2 h^2}{4 \pi^2 m^2 r_n^2}$
or $\quad \frac{Z e^2}{\epsilon_o}=\frac{n^2 h^2}{r_n \cdot \pi m}$ $\quad$ $\therefore r_n=\frac{\epsilon_0 n^2 h^2}{\pi m Z e^2}$
or $\quad \underline {r_n=\frac{\epsilon_0 h^2}{\pi m Z e^2} \cdot n^2}$ $\ldots (3)$
Speed of electron in stable orbit
$\because \quad v_n=\frac{n h}{2 \pi m r_n}$
Put the value of $r_n$ from eqn. (3)
$v_n=\frac{n h}{2 \pi m\left[\frac{\epsilon_0 h^2 n^2}{\pi m e^2 Z}\right]} \Rightarrow \underline {v_n=\frac{Z e^2}{2 \epsilon_0 h n}}$

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Question 35 Marks

On the basis of Bohr's postulates, derive the expression for orbital speed of electron for $n ^{\text {th }}$ stable orbit in hydrogen atom.
Energy in ground state of hydrogen atom is $(-) XeV$. What will be the kinetic energy in this state?

Answer

Bohr's first postulate :
The electrons revolves around the nucleus in different stable circular orbits of the atom.
The required contripetal force to more the electron in circular orbit provided by the Coloumb's force, works between the electron and charge of the nucleus.
Therefore, from the Bohr's first postulate,
$\frac{m v^2}{r} =\frac{KZ e^2}{r^2}$
$\frac{m v^2}{r} =\frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{r^2} $
$\Rightarrow m v^2 r=\frac{Z e^2}{4 \pi \epsilon_0} \ldots(1)$
Bohr's second postulate :
Electron revolves only there orbits in which angular momentum $(L =m v r)$ is multiple integer of $\frac{h}{2 \pi}$.
According to Bohr's this postulate:
$L=\frac{n h}{2 \pi} \text { or } m v r=n\left(\frac{h}{2 \pi}\right) \ldots(2) $
Where $h=$ Planck's constant and $n=1,2,3, \ldots$ are the principal quantum numbers.
From $(1)\ m v^2 r=\frac{Z e^2}{4 \pi \epsilon_0}$
or $ m v r . v=\frac{Z e^2}{4 \pi \epsilon_0} \ldots(3) $
Put the value of $m v r$ from eqn. $(2)$
$n\left(\frac{h}{2 \pi}\right) v=\frac{Z e^2}{4 \pi \epsilon_0}$
or $v=\frac{Z e^2}{4 \pi \epsilon_0} \times\left(\frac{2 \pi}{n h}\right) \text { or } v=\left(\frac{Z e^2}{2 \epsilon_0 h}\right) \frac{1}{n} \ldots (4) $
$($where, $n=1,2,3, \ldots)$
Equation $(4)$ is the general formula of velocity of electron in stable orbits. In this formula, $e, \epsilon_0$ and $h$ are the universal constants. $Z$ is constant specifically for atom therefore
$v \propto \frac{1}{n}$
That is, "The velocity of electron in stable orbits is inversely proportional to the orbit number i.e. principal quantum."
Value of kinetic energy of electron :
$K=-(-) XeV$
$K=XeV$

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Question 45 Marks

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Answer

Given that, energy of electron beam = 12.5 eV Energy of electron of hydrogen atom, in ground level = -13.6 eV
Therefore, energy of electron in hydrogen orbits by the bombardment of beam of electron
$=-13.6+12.5=-1.1 eV$
This energy of electron in $n=3$, is greater than
$E_3=-\frac{13.6}{9}=-1.5 eV$
Therefore electron will excited upto third level (orbit) by the bombardment of electron beam and transition possibly as per figure:
(i) From $n_2=3$ to $n_1=1$
(ii) From $n_2=2$ to $n_1=1$
In above will be Lyman series and in
(iii) From $n_3=3$ to $n_1=2$, Balmer series.

Again, since $h v=\frac{h c}{\lambda}=\Delta E$
Therefore wavelength of emitted radiation will be :
$\lambda=\frac{h c}{\Delta E (\text { in joule })}$
Therefore (i) transition from $n_3=3$ to $n_1=1$
$\lambda_1=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{[-1.5-(-13.6)] \times 1.6 \times 10^{-19}}$
$=\frac{19.89 \times 10^{-7}}{12.1 \times 1.6}=\frac{19.89 \times 10^{-7}}{19.36}$
$=1.027 \times 10^{-7}$ meter
$=102.7 \times 10^{-9}$ meter
or $\lambda_1=102.6 nm$.
(ii) For transition from $n_2=2$ to $n_1=1$
$\lambda_2=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{[-3.4-(-13.6)] \times 1.6 \times 10^{-19}}$
$=\frac{19.89 \times 10^{-7}}{10.2 \times 1.6}=\frac{19.89 \times 10^{-7}}{16.32}$ meter
$\begin{aligned}=1.219 \times 10^{-7} =121.9 \times 10^{-9} \text { meter } \\ =121.9 mm .\end{aligned}$
(iii) For transition, from $n_2=3$ to $n_1=2$
$\lambda_3=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{[-1.5-(-3.4)] \times 1.6 \times 10^{-19}}$
$=\frac{19.89 \times 10^{-7}}{1.9 \times 1.6}=\frac{19.89 \times 10^{-7}}{3.040}$ meter
$=6.543 \times 10^{-7}$ meter
$=654.3 \times 10^{-9}$ meter
$\lambda_3=654.3 nm$.

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Question 55 Marks

Using the Bohr's model, calculate the speed of the electron in a hydrogen atom in the n = 1 2, and 3 levels.

Answer

Using Bohr's Model, we know that radius and speed of electron in n^the orbit:
$r_n=\frac{n^2 h^2}{4 \pi^2 K m e^2}$ ....(1)
or $\quad v_n=\frac{n h}{2 \pi m r_n}=\frac{n h}{2 \pi m} \times \frac{4 \pi^2 K m e^2}{n^2 h^2}$
$\therefore \quad v_n=\frac{2 \pi K e^2}{n h}=\frac{1}{4 \pi \in_0} \frac{2 \pi e^2}{h} \frac{1}{n} \quad \because K=\frac{1}{4 \pi \in_0}$
$v_n=\frac{1}{4 \pi \in_0} \cdot \frac{2 \pi e^2}{h} \cdot \frac{1}{n}$ ....(2)
We know that,
$\begin{array}{l}e=1.6 \times 10^{-19} C \\ h=6.63 \times 10^{-34} J- s ,\end{array}$
$K=\frac{1}{4 \pi \in_0}=9 \times 10^9 Nm ^2 / C ^2$
All these values put in eqn. (2)
$v_n=\frac{9 \times 10^9 \times 2 \times 3.14 \times\left(1.6 \times 10^{-19}\right)^2}{6.63 \times 10^{-34}} \times \frac{1}{n}$
$v_n=\frac{21.871 \times 10^5}{n} m / s$
Let in $n=1,2$ and 3 , the speed of electron $v_1, v_2$ and $v_3$.
$v_1=\frac{21.871}{1} \times 10^5=21.871 \times 10^5 m / sec$
$\simeq 2.19 \times 10^6 m / sec$
$v_2=\frac{21.871}{2} \times 10^5 m / sec =10.935 \times 10^5 m / sec$
$=1.094 \times 10^6 m / sec$
and $v_3=\frac{21.871}{3} \times 10^5 m / sec$
$=7.290 \times 10^5 m / sec =0.729 \times 10^6 m / sec$

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