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In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number (n - 1). if n >> 1 , the frequency of radiation emitted is proportional to :

 

 

Option: 1

\frac{1}{n^{3}}


Option: 2

\frac{1}{n}


Option: 3

\frac{1}{n^{2}}


Option: 4

\frac{1}{n^{\frac{3}{2}}}


Answers (1)

best_answer

As we have learnt

\Delta E=h\nu=E_{0}.\left ( \frac{1}{\left ( n-1 \right )^{2}} -\frac{1}{n^{2}}\right )

or Frequency = \nu=\frac{E_{0}}{h }.\frac{n^{2}-\left ( n-1 \right )^{2}}{[n\left ( n-1 \right )]^{2}}

                       =\left ( \frac{E_{0}}{h } \right )\frac{2n-1}{n^2\left ( n-1 \right )^{2}}

Since n>>1 \Rightarrow Frequency \alpha 1/n^3

Posted by

HARSH KANKARIA

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