Problem

A photon is emitted from a hydrogen atom, which undergoes a transition from the $n = 3$ state to the $n = 2$ state. Calculate:
(a) The energy of the emitted photon,
(b) The wavelength of the photon, and
(c) The frequency of the photon.

Data

• Initial Energy Level ($n$): $3$
• Final Energy Level ($p$): $2$
• Rydberg Constant ($R_H$): $1.0974 \times 10^{7} \, \mathrm{m}^{-1}$
• Planck’s Constant ($h$): $6.626 \times 10^{-34} \, \mathrm{Js}$
• Speed of Light ($c$): $3 \times 10^{8} \, \mathrm{m/s}$

To Find:

• Wavelength ($\lambda$)
• Energy ($E$)
• Frequency ($f$)

Prerequisite Concepts

1. Wavelength Formula (Hydrogen Atom):

$$\frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$$

2. Energy of a Photon (Planck’s Hypothesis):

$$E = \frac{hc}{\lambda}$$

3. Frequency of a Photon:

$$f = \frac{E}{h}$$

4. Energy Conversion:
   Convert energy from joules to electronvolts using:

$$1 \, \mathrm{eV} = 1.602 \times 10^{-19} \, \mathrm{J}$$

Solution

Part (a): Calculate the Wavelength ($\lambda$)

Use the formula for the hydrogen atom’s wavelength:

$$\frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$$

Substitute $R_H = 1.0974 \times 10^{7} \, \mathrm{m}^{-1}$, $p = 2$, and $n = 3$:

$$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$$ $$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{1}{4} - \frac{1}{9} \right)$$ $$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{9 - 4}{36} \right)$$ $$\frac{1}{\lambda} = 1.0974 \times 10^{7} \cdot 0.138888$$ $$\frac{1}{\lambda} = 0.152417 \times 10^{7} \, \mathrm{m}^{-1}$$ $$\lambda = \frac{1}{0.152417 \times 10^{7}}$$ $$\lambda = 6.560974 \times 10^{-7} \, \mathrm{m}$$

Convert to nanometers:

$$\lambda = 656.09 \, \mathrm{nm}$$

Part (b): Calculate the Energy ($E$)

Using Planck’s hypothesis:

$$E = \frac{hc}{\lambda}$$

Substitute $h = 6.626 \times 10^{-34} \, \mathrm{Js}$, $c = 3 \times 10^{8} \, \mathrm{m/s}$, and $\lambda = 6.560974 \times 10^{-7} \, \mathrm{m}$:

$$E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^{8}}{6.560974 \times 10^{-7}}$$ $$E = 3.00298 \times 10^{-19} \, \mathrm{J}$$

Convert to electronvolts:

$$E = \frac{3.00298 \times 10^{-19}}{1.602 \times 10^{-19}}$$ $$E = 1.8912 \, \mathrm{eV}$$

Part (c): Calculate the Frequency ($f$)

Using Planck’s formula:

$$f = \frac{E}{h}$$

Substitute $E = 3.00298 \times 10^{-19} \, \mathrm{J}$ and $h = 6.626 \times 10^{-34} \, \mathrm{Js}$:

$$f = \frac{3.00298 \times 10^{-19}}{6.626 \times 10^{-34}}$$ $$f = 4.57 \times 10^{14} \, \mathrm{Hz}$$

Answer

(a) Wavelength:

$$\lambda = 656.09 \, \mathrm{nm}$$

(b) Energy:

$$E = 1.8912 \, \mathrm{eV}$$

(c) Frequency:

$$f = 4.57 \times 10^{14} \, \mathrm{Hz}$$