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Question

Physics Question on Atoms

A 12.5eV12.5\, 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

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

Explanation

Solution

E=hcλλ=hcE=6.62×1034×3×10812.5×1.6×1019E=\frac{h c}{\lambda} \Rightarrow \lambda=\frac{h c}{E}=\frac{6.62\times10^{-34}\times3\times10^{8}}{12.5\times1.6\times10^{-19}}
=993A?=993 A?
1λ=R(1n121n22)\frac{1}{\lambda}= R \left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2^{2}}}\right)
(where Rydberg constant ,R= 1.097 x 10710^{7})
or,1993×1010=1.097×107(1121n22)or, \, \frac{1}{993\times10^{-10}}=1.097\times10^{7} \left(\frac{1}{1^{2}}-\frac{1}{n_{2^{2}}}\right)
Solving we get n2n_{2} = 3
Spectral lines Total number of spectral lines = 3
Two lines in Lyman series for n1=1,n2=2n_{1}=1, n_{2} =2
and n1=1,n2=3n_{1}=1, n_{2} =3 and one in Balmer series
for n1=2,n2=3n_{1} =2, n_{2} =3