In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell the wavelength of emitted radiation is $\lambda$. If an electron jumps from N-shell to the L-shell the wavelength of emitted radiation will be:

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Modern Physics

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell the wavelength of emitted radiation is

λ

. If an electron jumps from N-shell to the L-shell the wavelength of emitted radiation will be:

Answer: (D)

The emitted wavenumber follows

\frac{1}{λ} = R_{H} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

. For the M-to-L jump (

3 \to 2

\frac{1}{λ} = R_{H} (\frac{1}{4} - \frac{1}{9}) = \frac{5 R _{H}}{36}

. For the N-to-L jump (

4 \to 2

\frac{1}{λ ^{'}} = R_{H} (\frac{1}{4} - \frac{1}{16}) = \frac{3 R _{H}}{16}

. Dividing,

\frac{λ ^{'}}{λ} = \frac{5/36}{3/16} = \frac{5}{36} \times \frac{16}{3} = \frac{20}{27}

, so

λ^{'} = \frac{20}{27} λ

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option