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JEE Main 20202 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
2 September
Shift
Evening Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2020 · 2 September · Evening Shift · Physics · Atomic Physics

In a hydrogen atom the electron makes a transition from (n + 1) ^...

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

Options

  1. A1n\dfrac{1}{n}
  2. B1n3\dfrac{1}{{n}^{3}}
  3. C1n2\dfrac{1}{{n}^{2}}
  4. D1n4\dfrac{1}{{n}^{4}}

Correct answer

B. 1n3\dfrac{1}{{n}^{3}}

Step-by-step solution

En=−Rhcn2{E}_{n} = - \dfrac{Rhc}{{n}^{2}}
En+1=−Rhc(n+1)2{E}_{n + 1} = - \dfrac{Rhc}{( n + 1{)}^{2}}
\DeltaE=En+1−En\DeltaE = {E}_{n + 1} - {E}_{n}
hv=Rhc[1n2−1(n+1)2]hv = Rhc \left[\dfrac{1}{{n}^{2}} - \dfrac{1}{( n + 1{)}^{2}}\right]
v=R.c[(n+1)2−n2n2(n+1)2]v = R . c\left[\dfrac{{\left(n + 1\right)}^{2} - {n}^{2}}{{n}^{2}{\left(n + 1\right)}^{2}}\right]
v=R.c[1+2nn2(n+1)2]v = R . c\left[\dfrac{1 + 2n}{{n}^{2}{\left(n + 1\right)}^{2}}\right]

n>>1⇒n > > 1 ⇒v=R.c[2nn2×n2]=2RCn3v = R . c\left[\dfrac{2n}{{n}^{2} \times {n}^{2}}\right] = \dfrac{2RC}{{n}^{3}}

v∞1n3v \infty \dfrac{1}{{n}^{3}}

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