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======= ####22. 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]

c3eec34ec6b1fad69db54a20ad4b2dca40c2aa54 (a) 3

(b) 4

(d) 2

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Answer:

Correct Answer: 22. (d)

Solution:

(d) For first line of Lyman series of hydrogen $\frac{h c}{λ_{1}} = R h c (\frac{1}{1^{2}} - \frac{1}{2^{2}})$

For second line of Balmer series of hydrogen like ion

$\frac{h c}{λ_{2}} = Z^{2} R h c (\frac{1}{2^{2}} - \frac{1}{4^{2}})$

By question, $λ_{1} = λ_{2}$

$\Rightarrow (\frac{1}{1} - \frac{1}{2}) = Z^{2} (\frac{1}{4} - \frac{1}{16})$ or $Z = 2$

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Atoms - Result Question 21

22. 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]

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22. 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]