Question

What are the (a) energy, (b) magnitude of the momentum, and (c) wavelength of the photon emitted when a hydrogen atom undergoes a transition from a state with n = 3 to a state with n = 1 ?

Short answer

1. The required energy is 12.1 eV .
2. The magnitude of the momentum is $6.46\times {10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s}$.
3. The wavelength of the photon is $1.02\times {10}^{-7}\mathrm{m}$.

Step-by-step solution

Describe the energy of the photon:

The Energy of the photons is given by,

$\mathbf{∆}\mathbf{E}\mathbf{=}\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{6}\left(\frac{1}{n_3^2}-\frac{1}{n_1^2}\right)$

Here, ${\mathbf{n}}_{\mathbf{1}}$ and ${\mathbf{n}}_{\mathbf{3}}$ are the energy levels.

(a) Define the energy of the photon:

Write the equation of the energy of the photons as below.

$∆\mathrm{E}=-13.6\left(\frac{1}{{\mathrm{n}}_3^2}-\frac{1}{{\mathrm{n}}_1^2}\right)$ ….. (1)

Substitute 3 for ${\mathrm{n}}_3$, and 1 for ${\mathrm{n}}_1$in equation (1).

$$\begin{gathered} ∆E=-13.6\left(\frac{1}{3^2}-\frac{1}{1^2}\right) \\ =-13.6\left(\frac{1}{9}-1\right) \\ =-13.6\left(-\frac{8}{9}\right) \\ =12.1\mathrm{eV} \end{gathered}$$

Therefore, the required energy is $12.1\mathrm{eV}$.

(b) Find the magnitude of the momentum:

The formula to calculate the magnitude of the momentum is given by,

$\mathrm{p}=\frac{∆\mathrm{E}}{\mathrm{c}}$ ….. (2)

Here, c is speed of the light.

Substitute $12.1\mathrm{eV}$for $∆\mathrm{E}$, and $3\times {10}^8\mathrm{m}/\mathrm{s}$ for c in equation (2).

$$\begin{gathered} p=\frac{12.1\mathrm{eV}}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ =\frac{\left(12.1\times 1.602\times {10}^{-19}\right)\mathrm{J}}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ =\frac{\left(19.38\times {10}^{-19}\right){\mathrm{kgm}}^2/{\mathrm{s}}^2}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ =6.46\times {10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the magnitude of the momentum is $6.46\times {10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s}$.

(c) Determine the wavelength of the photon:

The formula to calculate the wavelength is given by,

$\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ ….. (3)

Here, h is plank constant.

Substitute $6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}$ for h, and $6.46\times {10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s}$for p in equation (3).

$$\begin{gathered} \lambda =\frac{6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}}{6.46\times {10}^{-27}\mathrm{kg}·\mathrm{m}/\mathrm{s}} \\ =1.02\times {10}^{-7}\mathrm{m} \end{gathered}$$

Therefore, the wavelength of the photon is $1.02\times {10}^{-7}\mathrm{m}$.