First slide

Bohr Model of the Hydrogen atom.

Question

If an electron in a hydrogen atom jumps from the $3^{r d}$ orbit to the $2^{r d}$ orbit, it emits a photon of wavelength $λ$ . When it jumps from the $4^{r d}$ orbit to the $3^{r d}$ orbit, the corresponding wavelength of the photon will be

Easy

A

$\frac{16}{25} λ$
B

$\frac{9}{16} λ$
C

$\frac{20}{7} λ$
D

$\frac{20}{13} λ$

Solution

When electron jumps from higher orbit to lower orbit then, wavelength of emitted photon is given by,
$\frac{1}{λ} = R (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$
$so, \frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 R}{36}$
$\begin{array}{l} and \frac{1}{λ^{'}} = R (\frac{1}{3^{2}} - \frac{1}{4^{2}}) = \frac{7 R}{144} \\ ∴ λ^{'} = \frac{144}{7} \times \frac{5 λ}{36} = \frac{20 λ}{7} \end{array}$

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