Q146 / 180 · AIIMS 2013 · Physics

Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is

  • A
    $\frac{7}{29}$
  • B
    $\frac{9}{31}$
  • Correct answer
    C
    $\frac{5}{27}$
  • D
    $\frac{3}{23}$

Solution

The transition equation for Lyman series is given by $\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$ for largest wavelength, $n=2$. $\therefore \quad \frac{1}{\lambda_{\max }}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$ The transition equation for Balmer series is given by $\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$ for largest wavelength, $n = 3$ $\therefore \quad \frac{1}{\lambda_{\max }}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ Therefore, $\frac{\lambda_{L_{\max }}}{\lambda_{B_{\max }}}=\frac{\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{\left(\frac{1}{1^2}-\frac{1}{2^2}\right)}=\frac{\frac{5}{36}}{\frac{3}{4}}=\frac{5}{27}$

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

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