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When electron moves a transition from (n-1) to n states, the frequency of emitted radiation is related to n according to (n >>1)

Option: 1

\mathrm{\nu\ \alpha\ n^{-3}}


Option: 2

\mathrm{\nu\ \alpha\ n^{-2}}


Option: 3

\mathrm{\nu\ \alpha\ n^{3}}


Option: 4

\mathrm{\nu\ \alpha\ n^{-\frac{2}{3}}}


Answers (1)

best_answer

As we learn

Bohr’s Model for Hydrogen Atom/Hydrogen like atoms -

The frequency of radiation (\nu) absorbed or emitted when the electron jumps between two stationary states that differ in energy by ?E, is given by:

\nu= \Delta E/h= \left ( E_{2} -E_{1}\right )/h

E2 is a higher energy state and E1 is the lower energy state

-

 

 The frequency of emitted radiation is given by

\nu =Rcz^{2}\frac{1}{(n-1)^{2}}-\frac{1}{n^{2}}

\nu =Rcz^{2}(\frac{n^{2}-(n-1)^{2}}{(n(n-1))^{2}})

\nu =Rcz^{2}(\frac{2n - 1}{n^{2}\left ( n-1 \right )^{2}})

for large value of n

\nu =\frac{Rcz^{2}}{n^{3}}

\nu \propto n^{-3}

 

Posted by

Shailly goel

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