Question

1. the energy level diagram below is for hydrogen. use this diagram to answer the following questions.

a. an electron undergoes a photon and jumps from n=2 to n=4. determine the energy of the photon absorbed in kj/mol.
b. determine both the frequency and wavelength of the absorbed photon.

Explanation:

Step1: Define energy level values

From the diagram:
$E_2 = -3.40\ \text{eV}$, $E_4 = -0.85\ \text{eV}$

Step2: Calculate photon energy

Energy = $|E_{\text{final}} - E_{\text{initial}}|$
$\Delta E = |-0.85 - (-3.40)| = 2.55\ \text{eV}$
Convert to joules: $1\ \text{eV}=1.602\times10^{-19}\ \text{J}$
$\Delta E = 2.55 \times 1.602\times10^{-19} = 4.085\times10^{-19}\ \text{J}$

Step3: Find frequency of photon

Use $E=h
u$, $h=6.626\times10^{-34}\ \text{J·s}$
$
u = \frac{\Delta E}{h} = \frac{4.085\times10^{-19}}{6.626\times10^{-34}} \approx 6.17\times10^{14}\ \text{Hz}$

Step4: Find wavelength of photon

Use $c=\lambda
u$, $c=3\times10^8\ \text{m/s}$
$\lambda = \frac{c}{
u} = \frac{3\times10^8}{6.17\times10^{14}} \approx 4.86\times10^{-7}\ \text{m} = 486\ \text{nm}$

Answer:

a. The energy of the absorbed photon is $\boldsymbol{2.55\ \text{eV}}$ (or $\boldsymbol{4.09\times10^{-19}\ \text{J}}$)
b. The frequency of the absorbed photon is $\boldsymbol{6.17\times10^{14}\ \text{Hz}}$, and its wavelength is $\boldsymbol{486\ \text{nm}}$ (or $\boldsymbol{4.86\times10^{-7}\ \text{m}}$)