Name: Moseley’s Law
Uploaded: 2019-07-19
Duration: 7 min 40 s
Description: Moseley’s law is an empirical law concerning the characteristic x-rays that are emitted by atoms. Let's watch this video to learn about Moseley's law.

Question

Find the longest and shortest wavelength when a hydrogen atom in the ground state is excited by radiations of wavelength

975 \overset{°}{A}

.

Accepted Answer

Given,

Wavelength $λ = 975 \overset{°}{A}$

The energy of the radiation is,

$E = \frac{h c}{λ}$ , where $h$ is the Planck constant and $c$ is the speed of light.

$E = \frac{6.626 \times 10^{- 34} \times 3 \times 10^{4}}{975 \times 10^{- 10}} \times \frac{1}{1.6 \times 10^{- 19} }$
$E = 12.75 eV$

$E_{n} = - 13.6 + 12.75 eV = - 0.85 eV$

So, it will reach the fourth excited state.

Rydberg formula is given by,

$\frac{1}{λ} = R_{H} [\frac{1}{{n_{f}}^{2}} - \frac{1}{{n_{i}}^{2}}]$ , where $R_{H}$ is the Rydberg constant, $n_{f}$ is the final state, and $n_{i}$ is the initial state.

Now for the longest wavelength transition is $4 \to 3$ .

$\frac{1}{λ_{\max}} = R_{H} [\frac{1}{3^{2}} - \frac{1}{4^{2}}] \Rightarrow \frac{1}{λ_{\max}} = 1.097 \times 10^{7} \times \frac{7}{16 \times 9} \Rightarrow λ_{\max} = 18752.4 \overset{°}{A}$

Now for the shortest wavelength transition is $4 \to 1$ .

$\frac{1}{λ_{\min}} = R_{H} [\frac{1}{1^{2}} - \frac{1}{4^{2}}] \Rightarrow \frac{1}{λ_{\min}} = 1.097 \times 10^{7} \times \frac{15}{16} \Rightarrow λ_{\min} = 973 \overset{°}{A}$

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