JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 27)
A particle of mass m moves in a circular orbit in a
central potential field U(r) = U0r4. If Bohr's quantization conditions
are applied, radii of possible
orbitals rn vary with $${n^{{1 \over \alpha }}}$$, where $$\alpha$$ is ____________.
central potential field U(r) = U0r4. If Bohr's quantization conditions
are applied, radii of possible
orbitals rn vary with $${n^{{1 \over \alpha }}}$$, where $$\alpha$$ is ____________.
Answer
3
Explanation
$$\overrightarrow F = - {{d\overrightarrow u } \over {dr}}$$
$$ = - {d \over {dr}}({U_0}{r^4})$$
$$\overrightarrow F = - 4{U_0}{r^3}$$
$$ \because $$ $${{m{v^2}} \over r} = 4{U_0}{r^3}$$
$$m{v^2} = 4{U_0}{r^4}$$
Then $$v \propto {r^2}$$
$$ \because $$ $$mvr = {{nh} \over {2\pi }}$$
Then $${r^3}\propto\,n$$
$$r\,\propto \,{(n)^{{1 \over 3}}}$$
So the value of $$\alpha = 3$$
$$ = - {d \over {dr}}({U_0}{r^4})$$
$$\overrightarrow F = - 4{U_0}{r^3}$$
$$ \because $$ $${{m{v^2}} \over r} = 4{U_0}{r^3}$$
$$m{v^2} = 4{U_0}{r^4}$$
Then $$v \propto {r^2}$$
$$ \because $$ $$mvr = {{nh} \over {2\pi }}$$
Then $${r^3}\propto\,n$$
$$r\,\propto \,{(n)^{{1 \over 3}}}$$
So the value of $$\alpha = 3$$
Comments (0)
