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12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. It will emit : 

Option: 1

2 lines in the Lyman series and 1 line in the Balmer series 


Option: 2

3 lines in the Lyman series 


Option: 3

1 line in the Lyman series and 2 lines in the Balmer series 


Option: 4

3 lines in the Balmer series 


Answers (1)

best_answer

In ground state, the energy of gaseous hydrogen at room temperature =−13.6eV

When it is bombarded with a 12.5 eV electron beam, the energy becomes −13.6+12.5=−1.1eV

i.e \quad E_{n}=-1.1 \mathrm{ev}

and we know that E_{n}=\frac{-13.6}{n^{2}}

So comparing we get

\quad-1.1 \mathrm{ev}=\frac{-1.3 \cdot 6}{n^{2}} \mathrm{ev}

n^{2}=12 \cdot 4 \rightarrow n=\sqrt{12 \cdot 4}=3 \cdot 52<4

So n=3 is the final excitation state

\text { For } n=3, E=\frac{-13.6}{9}=-1.5 \mathrm{eV}

 It can be concluded that the electron has jumped from n=1 to n=3 level. 

On de-excitation, the electron may jump from n=3 to n=2 giving rise to the Balmer series.  It may also jump from n=3 to n=1, giving rise to the Lyman series,

It may also jump from n=2 to n=1, giving rise to the Lyman series,

So It will emit 2 lines in the Lyman series and 1 line in the Balmer series 

So the Correct option is A.

 

 

Posted by

Ritika Harsh

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