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PhysicsModern PhysicsAtomic structureHard2 minPYQ_2020

PhysicsHardsingle choice

In a hydrogen atom the electron makes a transition from ${(n + 1)}^{th}$ level to the $n^{th}$ level. If $n > > 1$ , the frequency of radiation emitted is proportional to :

Options:

Answer:

Solution:

$E_{n} = - \frac{R h c}{n^{2}}$
$E_{n + 1} = - \frac{R h c}{(n + 1)^{2}}$
$Δ E = E_{n + 1} - E_{n}$
$h v = Rhc [\frac{1}{n^{2}} - \frac{1}{(n + 1)^{2}}]$
$v = R . c [\frac{{(n + 1)}^{2} - n^{2}}{n^{2} {(n + 1)}^{2}}]$
$v = R . c [\frac{1 + 2 n}{n^{2} {(n + 1)}^{2}}]$

$n > > 1 \Rightarrow$ $v = R . c [\frac{2 n}{n^{2} \times n^{2}}] = \frac{2 R C}{n^{3}}$

$v \infty \frac{1}{n^{3}}$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2020

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