JEE Advance - Physics (2020 - Paper 1 Offline - No. 11)
A particle of mass m moves in circular orbits with potential energy V(r) = Fr, where F is a positive
constant and r is its distance from the origin. Its energies are calculated using the Bohr model. If the
radius of the particle’s orbit is denoted by R and its speed and energy are denoted by v and E,
respectively, then for the nth orbit (here h is the Planck’s constant)
$$R \propto {n^{{1 \over 3}}}$$ and $$v \propto {n^{{2 \over 3}}}$$
$$R \propto {n^{{2 \over 3}}}$$ and $$v \propto {n^{{1 \over 3}}}$$
$$E = {3 \over 2}{\left( {{{{n^2}{h^2}{F^2}} \over {4{\pi ^2}m}}} \right)^{{1 \over 3}}}$$
$$E = 2{\left( {{{{n^2}{h^2}{F^2}} \over {4{\pi ^2}m}}} \right)^{{1 \over 3}}}$$
Explanation
$$|Force| = \left| { - {{dV} \over {dr}}} \right| \Rightarrow F = {{m{v^2}} \over r}$$ .....(i)
$$mvr = {{nh} \over {2\pi }}$$ .....(ii)
From Eqs. (i) and (ii), we get
$$ \therefore $$ $$F = {m \over r}{\left( {{{nh} \over {2\pi mr}}} \right)^2}$$ $$ \because $$ $$\left( {v = {{nh} \over {2\pi mr}}} \right)$$
or $$F = {{m{n^2}{h^2}} \over {4{\pi ^2}{m^2}{r^3}}}$$
or $$F = {{{n^2}{h^2}} \over {4{\pi ^2}m{r^3}}}$$
$${r^3} \propto \,{n^2}$$ (as F = constant)
$$r \propto \,{n^{2/3}}$$
$$v = {{nh} \over {2\pi mr}}$$
$$v \propto {n \over {{n^{2/3}}}}$$
or $$v \propto {n^{1/3}}$$
$$ \therefore $$ $$KE = {1 \over 2}m{v^2} = {{Fr} \over 2}$$
$$ \because $$ $$TE = PE + KE = Fr + {{Fr} \over 2}$$
$$ \Rightarrow {{3Fr} \over 2} = TE$$
or $$TE = E = {{3F} \over 2} \times {\left( {{{{n^2}{h^2}} \over {4{\pi ^2}rnF}}} \right)^{1/3}}$$
$$ = {3 \over 2}{\left( {{{{n^2}{h^2}{F^2}} \over {4{\pi ^2}m}}} \right)^{1/3}}$$
$$mvr = {{nh} \over {2\pi }}$$ .....(ii)
From Eqs. (i) and (ii), we get
$$ \therefore $$ $$F = {m \over r}{\left( {{{nh} \over {2\pi mr}}} \right)^2}$$ $$ \because $$ $$\left( {v = {{nh} \over {2\pi mr}}} \right)$$
or $$F = {{m{n^2}{h^2}} \over {4{\pi ^2}{m^2}{r^3}}}$$
or $$F = {{{n^2}{h^2}} \over {4{\pi ^2}m{r^3}}}$$
$${r^3} \propto \,{n^2}$$ (as F = constant)
$$r \propto \,{n^{2/3}}$$
$$v = {{nh} \over {2\pi mr}}$$
$$v \propto {n \over {{n^{2/3}}}}$$
or $$v \propto {n^{1/3}}$$
$$ \therefore $$ $$KE = {1 \over 2}m{v^2} = {{Fr} \over 2}$$
$$ \because $$ $$TE = PE + KE = Fr + {{Fr} \over 2}$$
$$ \Rightarrow {{3Fr} \over 2} = TE$$
or $$TE = E = {{3F} \over 2} \times {\left( {{{{n^2}{h^2}} \over {4{\pi ^2}rnF}}} \right)^{1/3}}$$
$$ = {3 \over 2}{\left( {{{{n^2}{h^2}{F^2}} \over {4{\pi ^2}m}}} \right)^{1/3}}$$
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