Problem

An electron drops from the second energy level ($n = 2$) to the first energy level ($n = 1$) within an excited hydrogen atom.
(a) Determine the energy of the photon emitted.
(b) Calculate the frequency of the photon emitted.
(c) Calculate the wavelength of the photon emitted.

Data

• Initial Energy Level ($n$): $2$
• Final Energy Level ($p$): $1$
• Planck’s constant ($h$): $6.626 \times 10^{-34} \, \mathrm{Js}$
• Speed of light ($c$): $3 \times 10^{8} \, \mathrm{m/s}$
• Ground state energy ($E_0$): $2.17 \times 10^{-18} \, \mathrm{J} = 13.6 \, \mathrm{eV}$

Prerequisite Concepts

1. Energy Difference ($E$):
   The energy of a photon emitted is equal to the difference in energy levels:

$$E = E_p - E_n$$

Where:

• $E_p = -\frac{E_0}{p^2}$
• $E_n = -\frac{E_0}{n^2}$

2. Frequency ($f$):
   By Planck’s quantum theory:

$$E = hf \quad \Rightarrow \quad f = \frac{E}{h}$$

3. Wavelength ($\lambda$):
   Using Planck’s hypothesis:

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

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

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

Solution

Part (a): Energy of the Photon ($E$)

Calculate $E_p$ and $E_n$:

$$E_p = -\frac{E_0}{p^2} = -\frac{13.6 \, \mathrm{eV}}{1^2} = -13.6 \, \mathrm{eV}$$ $$E_n = -\frac{E_0}{n^2} = -\frac{13.6 \, \mathrm{eV}}{2^2} = -3.4 \, \mathrm{eV}$$

Energy difference:

$$E = E_p - E_n = -13.6 \, \mathrm{eV} - (-3.4 \, \mathrm{eV})$$ $$E = 10.2 \, \mathrm{eV}$$

Convert to joules:

$$E = 10.2 \, \mathrm{eV} \times 1.602 \times 10^{-19} \, \mathrm{J/eV}$$ $$E = 1.63404 \times 10^{-18} \, \mathrm{J}$$

Part (b): Frequency of the Photon ($f$)

Using Planck’s equation:

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

Substitute $E = 1.63404 \times 10^{-18} \, \mathrm{J}$ and $h = 6.626 \times 10^{-34} \, \mathrm{Js}$:

$$f = \frac{1.63404 \times 10^{-18}}{6.626 \times 10^{-34}}$$ $$f = 2.466 \times 10^{15} \, \mathrm{Hz}$$

Part (c): Wavelength of the Photon ($\lambda$)

Using the wavelength formula:

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

Substitute $h = 6.626 \times 10^{-34} \, \mathrm{Js}$, $c = 3 \times 10^{8} \, \mathrm{m/s}$, and $E = 1.63404 \times 10^{-18} \, \mathrm{J}$:

$$\lambda = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^{8}}{1.63404 \times 10^{-18}}$$ $$\lambda = 1.216 \times 10^{-7} \, \mathrm{m}$$

Convert to nanometers:

$$\lambda = 121.6 \, \mathrm{nm}$$

Answer

(a) Energy of the photon:

$$E = 10.2 \, \mathrm{eV} \, \text{or} \, 1.63404 \times 10^{-18} \, \mathrm{J}$$

(b) Frequency of the photon:

$$f = 2.466 \times 10^{15} \, \mathrm{Hz}$$

(c) Wavelength of the photon:

$$\lambda = 121.6 \, \mathrm{nm}$$

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