Q156 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Physics

The shortest wavelength in Balmer series of hydrogen atom spectrum is approximately equal to (use $\mathrm{R}_{\mathrm{H}}=1.097 \times 10^7 \mathrm{~m}^{-1}$ )

  • Correct answer
    A
    $3646 Å$
  • B
    $912 Å$
  • C
    $364.6 Å$
  • D
    $91.2 Å$

Solution

For shortest wavelength $\mathrm{n}_{\mathrm{f}}=2$ and $\mathrm{n}_{\mathrm{i}}=\infty$ $$ \begin{aligned} & \text { So, } \frac{1}{\lambda_{\text {short }}}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}_{\mathrm{f}}^2}-\frac{1}{\mathrm{n}_{\mathrm{i}}^2}\right) \\ & \Rightarrow \frac{1}{\lambda_{\mathrm{s}}}=1.097 \times 10^7\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right) \\ & \Rightarrow \frac{1}{\lambda_{\mathrm{s}}}=2742500 \mathrm{~m}^{-1} \\ & \Rightarrow \lambda_{\mathrm{s}}=3646 \times 10^{-10} \mathrm{~m} \end{aligned} $$

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

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