Q156 / 160 · AP EAMCET 2024 (22 May Shift 2) · Physics

The principle quantum number ' $n$ ' corresponding to the exited state of $\mathrm{He}^{+}$ion, if on transition to the ground state two photons in succession with wavelength $1026 \mathrm{~A}^{\circ}$ and $304 \mathrm{~A}^{\circ}$ are emitted ( $\mathrm{R}=1.097 \times 10^7 \mathrm{~m}^{-1}$ )

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

Solution

For first photon, $\begin{aligned} & \lambda=304 Å, \mathrm{n}_1=1 \\ & \frac{1}{\lambda}=\mathrm{Rz}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\end{aligned}$ $\begin{aligned} & \Rightarrow \frac{1}{304 \times 10^{-10}}=1.097 \times 10^7 \times(2)^2\left(\frac{1}{1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\ & \therefore \quad \mathrm{n}_2=2 \end{aligned}$ For second photon, $\begin{aligned} & \lambda=1026 Å, \mathrm{n}_1=2, \mathrm{n}_2=\mathrm{n} \\ & \frac{1}{1026 \times 10^{-10}}=1.097 \times 10^7 \times(2)^2\left(\frac{1}{2^2}-\frac{1}{\mathrm{n}^2}\right) \\ & \therefore \mathrm{n}=6 \end{aligned}$

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

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