"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 ____________."

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Accepted Answer

"$\overrightarrow F = - {{d\overrightarrow u } \over {dr}}$

$= - {d \over {dr}}({U_0}{r^4})$

$\overrightarrow F = - 4{U_0}{r^3}$

$\because$ ${{m{v^2}} \over r} = 4{U_0}{r^3}$

$m{v^2} = 4{U_0}{r^4}$

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 $\alpha = 3$"

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 r_n vary with ${n^{{1 \over \alpha }}}$, where $\alpha$ is ____________.

Solution

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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 ____________.

Solution

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More Atoms and Nuclei questions

Drill 25 more like these. Every day. Free.

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 r_n vary with ${n^{{1 \over \alpha }}}$, where $\alpha$ is ____________.