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The energy absorbed in the transition of the electron from n=2 to 1 in H-atom will be 

  • Option 1)

    10.2 meV

  • Option 2)

    3.4 meV

  • Option 3)

    13.6 meV

  • Option 4)

    12.75 meV

 

Answers (1)

best_answer

As we learnt in 

Lyman Series spectrum -

 \frac{1}{\lambda }= RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

Where n_{1}=1\ and\ n_{2}=2, 3, 4.....

This lies in Ultraviolet region

-

 

 \frac{1}{X}=\:RZ^{2}(\frac{1}{n^{2}}-\frac{1}{n{_{2}}^{2}})

where n_{1}=1 for Lyman series

n_{2}=2 (given)

Z=1 (hydrogen)

Rydberg constant, R = 1.097\times10^{7}\:m^{-1}

\frac{1}{\lambda }\:=\:1.097\times10^{7}\times 1 \times (1- \frac{1}{4})

\frac{1}{\lambda }\:=\: \frac{1}{1.212 \times 10^{-7}\:m}

Energy absorbed = \frac{hc}{\lambda }

=        6.626 \times\: \frac{10^{-34}\times 3 \times 10^{8}}{1.212 \times 10^{-7} \times 1.6 \times 10^{-16}}\:eV= 10.2 meV


Option 1)

10.2 meV

Option 2)

3.4 meV

Option 3)

13.6 meV

Option 4)

12.75 meV

Posted by

Plabita

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