Q149 / 180 · COMEDK 2023 (Memory Based) · Physics

The wavelength of the first line of Lyman series for $\mathrm{H}$ - atom is equal to that of the second line of Balmer series for a $\mathrm{H}$-like ion. The atomic number $\mathrm{Z}$ of $\mathrm{H}$-like ion is

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

Solution

Lyman series of $\mathrm{H}$-atom, we can write $\frac{h c}{\lambda}=R h c\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$ where, symbols have their usual meaning and for second line of Balmer series of $\mathrm{H}$-like ion $\frac{h c}{\lambda}=Z^2 R h c\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \text {. }$ Therefore, $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=Z^2\left(\frac{1}{4}-\frac{1}{16}\right)$ $\begin{aligned} \left(1-\frac{1}{4}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\ Z & =2.19 \simeq 2 \end{aligned}$ The approximately value of $Z$ will be 2 .

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

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