Q107 / 150 · MHT CET 2023 (10 May Shift 1) · Physics

The shortest wavelength in the Balmer series of hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen like atom of atomic number $z$. The value of $\mathrm{z}$ is

  • Correct answer
    A
    $2$
  • B
    $3$
  • C
    $4$
  • D
    $6$

Solution

Using Rydberg's formula, $\frac{1}{\lambda}=R_H Z^2\left[\frac{1}{\mathrm{n}^2}-\frac{1}{\mathrm{~m}^2}\right]$ where $R_H$ is the Rydberg's constant For calculating the shortest wavelength in the Balmer series of hydrogen atom, $\begin{aligned} & \mathrm{n}=2, \mathrm{~m}=\infty \text { and } \mathrm{Z}=1 \\ & \therefore \quad \lambda_1=\frac{4}{\mathrm{R}_{\mathrm{H}}} \end{aligned}$ For hydrogen like atom, the shortest wavelength is given by, In Brackett series $\begin{array}{ll} & \mathrm{n}=4, \mathrm{~m}=\infty \\ \therefore \quad & \frac{1}{\lambda_2}=\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2\left(\frac{1}{16}\right) \\ \therefore \quad & \lambda_2=\frac{16}{\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2} \\ & \text { Given: } \lambda_1=\lambda_2, \\ \therefore \quad & \frac{4}{\mathrm{R}_{\mathrm{H}}}=\frac{16}{\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2} \Rightarrow \mathrm{Z}^2=4 \\ \therefore \quad & \mathrm{Z}=2 \end{array}$

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

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