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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2020

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A Bohr hydrogen atom undergoes a transition $n = 5 \to n = 4$ and emits a photon of frequency $υ$ . Frequency of circular motion of electron in $n = 4$ orbit is $v_{4}$ . The ratio $\frac{ν}{ν_{4}}$ is

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Answer:

Solution:

$E_{n} = - \frac{m z^{2} e^{4}}{8 ε_{0}^{2} n^{2} h^{2}}$

So, $h v = + \frac{m z^{2} e^{4}}{8 ε_{0}^{2} h^{2}} [\frac{1}{16} - \frac{1}{25}]$

$∴$ $v = \frac{m z^{2} e^{4}}{8 ε_{0}^{2} h^{3}} [\frac{9}{16 \times 25}]$ $. . . (1)$

And frequency $v_{4} = \frac{1}{T} = \frac{v}{2 π r}$

$= (\frac{Z e^{2}}{2 ε_{0} n h}) \frac{1}{2 π} (\frac{π m z e^{2}}{ε_{0} h^{2} n^{2}}) = \frac{z^{2} e^{4} m}{4 ε_{0}^{2} n^{3} h^{3}}$ $. . . (2)$

$∴ \frac{v}{v_{4}} = \frac{18}{25} = 0.72$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2020

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