Question: Q. 7. In a Geiger-Marsden experiment, calculate the distance of the closest approach to the nucleus of $Z=80$, when an $\alpha$-particle of $8 \mathrm{MeV}$ energy impinges on it before it comes momentarily to rest and reverses its direction.
How will the distance of the closest approach be affected when the kinetic energy of the $\alpha$-particle is doubled?
A [OD I, II, III 2012]
$$ \begin{aligned} \frac{(\mathrm{Ze})(2 e)}{4 \pi \varepsilon_{0}\left(r_{0}\right)} & =E \ r_{0} & =\frac{2 Z e^{2}}{4 \pi \varepsilon_{0}(E)}=\frac{2 Z e^{2}}{4 \pi \varepsilon_{0} \times E} \ r_{0} & =\frac{9 \times 10^{9} \times 2 \times 80 \times\left(1.6 \times 10^{-19}\right)^{2}}{8 \times 10^{6} \times\left(1.6 \times 10^{-19}\right)} \mathrm{m}^{1 / 2} \ r_{0} & =\frac{18 \times 1.6 \times 10^{-10} \times 80}{8 \times 10^{6}} \quad 1 / 2 \ r_{0} & =2.88 \times 10^{-14} \mathrm{~m} \ r_{0} & \propto \frac{1}{\mathrm{~K} . \mathrm{E} .} \end{aligned} $$
If K.E. becomes twice then $r_{0}{ }^{\prime}=\frac{r_{0}}{2}$,
i.e. the distance of closest approach becomes half. $1 / 2$
[CBSE Marking Scheme 2012]
Detailed Answer :
At closest approach, all K.E. of $\alpha$-particle is totally converted into P.E. and it is the total energy of $\alpha$ - particle.
K.E. $=8 \mathrm{MeV}=8 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$.
It is converted into Potential energy.
$$ \begin{gathered} \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Z e 2 e}{r_{0}}=12.8 \times 10^{-13} \ 9 \times 10^{9} \times \frac{80 \times 2 \times 1.6 \times 10^{-19}}{r_{0}}=12.8 \times 10^{-13} \ r_{0}=\frac{9 \times 10^{9} \times 160 \times\left(1.6 \times 10^{-19}\right)^{2}}{12.8 \times 10^{-13}}(\mathrm{~m}) \quad 1 / 2 \end{gathered} $$
$$ \begin{aligned} \qquad r_{0} & =\frac{9 \times 1.6 \times 1.6 \times 1.6 \times 10^{-14}}{12.8} \ & =2.88 \times 10^{-14} \mathrm{~m} \ \text { As we see that K.E. } & =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Ze} 2 e}{r_{0}} \ \text { It shows } \quad \text { K.E. } & \propto \frac{1}{r_{0}} \end{aligned} $$
if K.E. becomes twice, closest approach would be half of original closest approach.
Hence, $\quad r_{0}{ }^{\prime}=\frac{r_{0}}{2}$
Q. 8. (i) Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but same orbital angular momentum according to the Bohr model? Justify your answer.
(ii) If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when $R$ $=10 \AA$.
A&E C [SQP 2016]
Show Answer
Solution:
Ans. (i) In absence of magnetic field, the energy is determined by the principle quantum number $n$, while $l$ is the orbital quantum number. If an electron is in $n^{\text {th }}$ state then the magnitude of the angular momentum is
$$ \frac{h}{2 \pi} l(l+1) $$
where, $l=0,1,2, \ldots \ldots \ldots . .,(n-1)$
Since $l=0,1,2, \ldots . .,(n-1)$, different values of $l$ are compatible with the same value of $n$. For example, when $n=3$, the possible values of $\operatorname{tare} 0,1,2$, and when $n=4$, the possible value of $h$ are $0,1,2,3$. Thus, the electron in one of the atoms could have $n=3, l=2$, while the electron in the other atom could have $n=4, l=4$. Therefore, according to quantum mechanics, it is possible for the electrons to have different energies but have the same orbital angular momentum.
1
(ii) For a point nucleus in $\mathrm{H}$-atom :
Ground state : $m v r=h, \frac{m v^{2}}{r_{B}}=-\frac{e^{2}}{r_{B}^{2}} \cdot \frac{1}{4 \pi \varepsilon_{0}}$
$\therefore \quad m \frac{h^{2}}{m^{2} r_{B}{ }^{2}} \cdot \frac{1}{r_{B}}=+\left(\frac{e^{2}}{4 \pi \varepsilon_{0}}\right) \frac{1}{r_{B}^{2}}$
$\therefore \quad\left(\frac{h^{2}}{m} \cdot \frac{4 \pi \varepsilon_{0}}{e^{2}}\right)=r_{B}=0.51 \AA$
If $R»r_{B}$ : the electron moves inside the sphere with radius $r_{B}^{\prime}\left(r_{B}^{\prime}=\right.$ new Bohr radius).
$$ \begin{aligned} & \text { Charge inside } r_{B}^{\prime}{ }^{4}=e\left(\frac{r_{B}^{\prime}{ }^{3}}{R^{3}}\right) \ & \therefore \quad r_{B}^{\prime}=\frac{h^{2}}{m}\left(\frac{4 \pi \varepsilon_{0}}{e^{3}}\right) \frac{R^{3}}{r_{B}^{\prime}{ }^{3}} \ & r_{B}^{\prime A}=(0.51 \AA) \cdot R^{3} . \quad \mathrm{R}=10 \AA \ & =510(\AA)^{4} \ & \therefore \quad r_{B}^{\prime}=(510)^{1 / 4} \AA<\mathrm{R} . \ & \text { K.E. }=\frac{1}{2} m v^{2}=\frac{m}{2} \cdot \frac{h}{m^{2} r_{B}^{\prime}{ }^{2}}=\frac{h^{2}}{2 m} \cdot \frac{1}{{r^{\prime}}{B}{ }^{2}} \ & =\left(\frac{h^{2}}{2 m r{B}^{2}}\right) \cdot\left(\frac{r_{B}^{2}}{r_{B}^{\prime 2}}\right) \ & =(13.6 \mathrm{eV}) \frac{(0.51)^{2}}{(510)^{1 / 2}} \ & \frac{3.54}{22.6}=0.16 \mathrm{eV} \ & \mathrm{E} .=+\left(\frac{e^{2}}{4 \pi \varepsilon_{0}}\right) \cdot\left(\frac{r_{\mathrm{B}}^{\prime}{ }^{2}-3 R^{2}}{2 R^{3}}\right) \ & =+\left(\frac{e^{2}}{4 \pi \varepsilon_{0}} \cdot \frac{1}{r_{B}}\right) \cdot\left(\frac{r_{B}^{1 / 2}-3 R^{2}}{R^{3}}\right) \ & =(27.2 \mathrm{eV})\left[\frac{0.51(\sqrt{510}-300)}{1000}\right] \ & =+(27.2 \mathrm{eV}) \cdot \frac{-141}{1000} \ & =-3.83 \mathrm{eV} \end{aligned} $$
[CBSE Marking Scheme, 2016]
