Wavelength of first line of Lyman series for hydrogen atom is equal to that of second line of Balmer

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12.Atoms

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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

A

$3$

B

$4$

C

$1$

D

$2$

Solution

For first line of Lyman series of hydrogen

$\frac{\mathrm{hc}} {\lambda_{1}}=\mathrm{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$

For second line of Balmer series of hydrogen like ion

$\frac{\mathrm{hc}} {\lambda_{2}}=\mathrm{Z}^{2} \mathrm{Rhc}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$

By question, $\lambda_{1}=\lambda_{2}$

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

Standard 12

Physics

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