Q131 / 200 · NEET 2011 (Screening) · Physics

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
    4
  • B
    1
  • Correct answer
    C
    2
  • D
    3

Solution

Lyman series for $\mathrm{H}$-ion $$ \frac{h c}{\lambda}=R h c\left(\frac{1}{1^2}-\frac{1}{2^2}\right) $$ And for H-like ion $$ \begin{aligned} \frac{h c}{\lambda} & =Z^2 R h c\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\ \therefore \quad\left(\frac{1}{1^2}-\frac{1}{2^2}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\ \left(1-\frac{1}{4}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\ Z & =2 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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