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  • A hydrogen atom initially in its ground state absorbs a photon and is in the excited state with energy 12.5 eV. Calculate the longest wavelength of the radiation emitted and identify the series to which it belongs. [Take Rydberg constant <img alt="R=1.

A hydrogen atom initially in its ground state absorbs a photon and is in the excited state with energy 12.5 eV. Calculate the longest wavelength of the radiation emitted and identify the series to which it belongs. [Take Rydberg constant R=1.1\times 10^{7} m^{-1}]

 

 

 

 
 
 
 
 

Answers (1)

The energy in the ground state is -13.6 \; eV

The energy of the photon =12.5 \; eV

The energy in the excited state

=-13.6+12.5

=1.1 \; eV

1.1 =\frac{13.6}{n^{2}}

n^{2}=\frac{13.6}{1.1}\approx 12.36

n\approx 3

\frac{1}{\lambda }=R\left ( \frac{1}{n_{f}^{2}}-\frac{1}{n;^{2}} \right )

For \lambda to be maximum n_{f}=2

\frac{1}{\lambda _{max}}=R\left ( \frac{1}{2^{2}}-\frac{1}{3^{2}} \right )

\lambda _{max}=6.55\times 10^{-7}m

and it belongs to Balmer Series.

Posted by

Safeer PP

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