Q111 / 125 · BITSAT 2016 · Physics

If the series limit wavelength of Lyman series for the hydrogen atom is $912 Å,$ then the series limit wavelength for Balmer series of hydrogen atoms is

  • A
    $912 Å$
  • B
    $912 \times 2 Å$
  • Correct answer
    C
    $912 \times 4 Å$
  • D
    $\frac{912}{2} Å$

Solution

$\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$ For limiting wavelength of Lyman series $$ n_{1}=1, n_{2}=\infty \quad \frac{1}{\lambda_{L}}=R $$ For limiting wavelength of Balmer series $n_{1}=2, n_{2}=\infty$ $$ \begin{array}{l} \frac{1}{\lambda_{B}}=R\left(\frac{1}{4}\right) \Rightarrow \lambda_{B}=\frac{4}{R} \\ \therefore \lambda_{B}=4 \lambda_{L}=4 \times 912 \AA \end{array} $$

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

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