JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 24)
A particle of mass m moves in a circular orbit in a central potential field U(r) = $${1 \over 2}$$ kr2. If Bohr 's
quantization conditions are applied, radii of possible orbitls and energy levels vary with quantum number n as :
rn $$ \propto $$ $$\sqrt n $$, En $$ \propto $$ n
rn $$ \propto $$ $$\sqrt n $$, En $$ \propto $$ $${1 \over n}$$
rn $$ \propto $$ n, En $$ \propto $$ n
rn $$ \propto $$ n2, En $$ \propto $$ $${1 \over {{n^2}}}$$
Explanation
Force due to this field, F = $$ - {{\partial U} \over {\partial r}}$$
F = $$ - {\partial \over {\partial r}}\left( {{1 \over 2}k{r^2}} \right)$$ = -kr
For circular orbit, $${{m{v^2}} \over r}$$ = -kr
$$ \Rightarrow $$ v $$ \propto $$ r ..... (1)
Fron Bohr’s quantization condition
mvr = $${{nh} \over {2\pi }}$$ .....(2)
From (1) and (2),
$${r_n}$$ $$ \propto $$ $${n^{{1 \over 2}}}$$
Given, U(r) = $${1 \over 2}$$kr2
$$ \Rightarrow $$ En = $$ - {1 \over 2}U\left( r \right)$$ = $$ - {1 \over 4}k{r^2}$$
$$ \Rightarrow $$ En $$ \propto $$ n
F = $$ - {\partial \over {\partial r}}\left( {{1 \over 2}k{r^2}} \right)$$ = -kr
For circular orbit, $${{m{v^2}} \over r}$$ = -kr
$$ \Rightarrow $$ v $$ \propto $$ r ..... (1)
Fron Bohr’s quantization condition
mvr = $${{nh} \over {2\pi }}$$ .....(2)
From (1) and (2),
$${r_n}$$ $$ \propto $$ $${n^{{1 \over 2}}}$$
Given, U(r) = $${1 \over 2}$$kr2
$$ \Rightarrow $$ En = $$ - {1 \over 2}U\left( r \right)$$ = $$ - {1 \over 4}k{r^2}$$
$$ \Rightarrow $$ En $$ \propto $$ n
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