Q85 / 100 · BITSAT 2024 (Memory Based Paper 2) · Physics

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is:

  • Correct answer
    A
    $4: 1$
  • B
    $1: 2$
  • C
    $1: 4$
  • D
    $2: 1$

Solution

Wavelength of H -atom is $\frac{1}{\lambda}=\mathrm{Rz}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$ Shortest wavelength for Balmer series: $\frac{1}{\lambda_B}=\mathrm{Rz}^2\left(\frac{1}{2^2}-\frac{1}{\infty}\right)$ Shortest wavelength for Lyman series: $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{Rz}^2\left(\frac{1}{1^2}-\frac{1}{\infty}\right)$ ....(ii) Dividing eq. (i) and (ii), $\lambda_{\mathrm{A}}: \lambda_1=4: 1$

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

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