Question

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

• A. 2 lines in the Lyman series and 1 line in the Balmar series
• B. 3 lines in the Lyman series
• C. 1 line in the Lyman series and 2 lines in the Balmar series
• D. 3 lines in the Balmer series

Answer 1

The Correct Option is A

$E=\frac{h c}{\lambda} \Rightarrow \lambda=\frac{h c}{E}=\frac{6.62\times10^{-34}\times3\times10^{8}}{12.5\times1.6\times10^{-19}}$
$=993 A?$
$\frac{1}{\lambda}= R \left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2^{2}}}\right)$
(where Rydberg constant ,R= 1.097 x $10^{7}$)
$or, \, \frac{1}{993\times10^{-10}}=1.097\times10^{7} \left(\frac{1}{1^{2}}-\frac{1}{n_{2^{2}}}\right)$
Solving we get $n_{2}$ = 3
Spectral lines Total number of spectral lines = 3
Two lines in Lyman series for $n_{1}=1, n_{2} =2$
and $n_{1}=1, n_{2} =3$ and one in Balmer series
for $n_{1} =2, n_{2} =3$