Atoms Ques 16
16. The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number $Z$ of hydrogen like ion is [2011]
(a) 3
(b) 4
(c) 1
(d) 2
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Answer:
Correct Answer: 16.(d)
Solution:
- (d) For first line of Lyman series of hydrogen $\frac{hc}{\lambda_1}=Rhc(\frac{1}{1^{2}}-\frac{1}{2^{2}})$
For second line of Balmer series of hydrogen like ion
$\frac{hc}{\lambda_2}=Z^{2} Rhc(\frac{1}{2^{2}}-\frac{1}{4^{2}})$
By question, $\lambda_1=\lambda_2$
$\Rightarrow(\frac{1}{1}-\frac{1}{2})=Z^{2}(\frac{1}{4}-\frac{1}{16})$ or $Z=2$
BETA
