Atoms & Nuclei

JEE Main level, single correct, medium

136 of 239

Question 136 of 239

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\gt \gt 1, the frequency of radiation emitted is proportional to :

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Answer: (B) 1n3\dfrac{1}{{n}^{3}}

En=−Rhcn2{E}_{n}=-\dfrac{Rhc}{{n}^{2}}
En+1=−Rhc(n+1)2{E}_{n+1}=-\dfrac{Rhc}{(n+1{)}^{2}}
ΔE=En+1−En\Delta E={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\gt \gt 1\Rightarrowv=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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