Question

In our world, the ionization potential energy of a hydrogen atom is $13.6eV$. On the other planet, this ionization potential energy will be :

• A. $13.6eV$
• B. $3.4eV$
• C. $1.5eV$
• D. $0.85eV$

Answer 1

The correct option is B $3.4eV$

For orbit $n$:
$m{v}_n{R}_n=2n\frac{h}{2\pi }\Rightarrow v_n{R}_n=\frac{nh}{m\pi }$...(i)

Electrostatic force is providing centripetal force for electron's circular motion:
$\frac{1}{4\pi {ϵ}_{\circ }}\frac{e^2}{R_{\circ }{}^2}=\frac{m{v}_n^2}{R_n}$ .....(ii)
$\frac{1}{4\pi {ϵ}_{\circ }}\frac{e^2}{m}=v_n^2{R}_n$...(iii)

Substituting $v_n{R}_n$ from equation (i)
$v_n\frac{nh}{m\pi }=\frac{1}{4\pi {ϵ}_{\circ }}\frac{e^2}{m}$
$v_n=\frac{e^2}{4{ϵ}_{\circ }nh}$

From relation (i) $R_n=\frac{nh}{m\pi v_n}$
$R_n=\frac{n{h}^{2}4{ϵ}_{\circ }}{m{e}^2\pi }$

Now the total energy will be :
$K.E.+P.E.$$=\frac{1}{2}m{v}_n^2-\frac{K{e}^2}{R_n}=\frac{1}{2}m(\frac{e^2}{4{ϵ}_{\circ }nh}{)}^2-\frac{e^2}{4\pi {ϵ}_{\circ }\frac{n{h}^{2}4{ϵ}_{\circ }}{m{e}^2\pi }}$
Substituting $n=1$ for ionization energy.
$=\frac{1}{2}m{v}_n^2-\frac{K{e}^2}{R_n}=\frac{1}{2}m(\frac{e^2}{4{ϵ}_{\circ }nh}{)}^2-\frac{K{e}^2}{\frac{n{h}^{2}4{ϵ}_{\circ }}{m{e}^2\pi }}$
$T.E{.}_{otherplanet}=\frac{m{e}^4}{32{ϵ}_{\circ }{}^2{h}^2}$

Similarly $T.E$ of hydrogen atom in earth can be calculated:
For orbit $n$:
$m{v}_n{R}_n=n\frac{h}{2\pi }\Rightarrow v_n{R}_n=\frac{nh}{2m\pi }$...(i)

Electrostatic force is providing centripetal force for electron's circular motion:
$\frac{1}{4\pi {ϵ}_{\circ }}\frac{e^2}{R_{\circ }{}^2}=\frac{m{v}_n^2}{R_n}$...(ii)
$\frac{1}{4\pi {ϵ}_{\circ }}\frac{e^2}{m}=v_n^2{R}_n$...(iii)

Substituting $v_n{R}_n$ from equation (i)
$v_n\frac{nh}{2m\pi }=\frac{1}{4\pi {ϵ}_{\circ }}\frac{e^2}{m}$
$v_n=\frac{e^2}{2{ϵ}_{\circ }nh}$

From relation (i) $R_n=\frac{nh}{2m\pi v_n}$
$R_n=\frac{n{h}^2{ϵ}_{\circ }}{m{e}^2\pi }$

Now the total energy will be :
$K.E.+P.E$ $=\frac{1}{2}m{v}_n^2-\frac{K{e}^2}{R_n}=\frac{1}{2}m(\frac{e^2}{2{ϵ}_{\circ }nh}{)}^2-\frac{e^2}{4\pi {ϵ}_{\circ }\frac{n{h}^2{ϵ}_{\circ }}{m{e}^2\pi }}$
substituting $n=1$ for ionization energy

$=\frac{1}{2}m{v}_n^2-\frac{K{e}^2}{R_n}=\frac{1}{2}m(\frac{e^2}{2{ϵ}_{\circ }nh}{)}^2-\frac{e^2}{4\pi {ϵ}_{\circ }\frac{n{h}^2{ϵ}_{\circ }}{m{e}^2\pi }}$
$T.E{.}_{earth}=-\frac{m{e}^4}{8{ϵ}_{\circ }{}^2{h}^2}$
Clearly $T.E{.}_{earth}=4T.E{.}_{otherplanet}$
Hence, $T.E{.}_{otherplanet}=\frac{T.E{.}_{earth}}{4}=\frac{-13.4}{4}=-3.4eV$