SOLUTION: Electromagnetic Spectrum Worksheet - Studypool - Free Printable

Problem Statement:

The problem involves calculating the wavelength of a photon emitted when an electron transitions from the \( n = 4 \) energy level to the \( n = 2 \) energy level in a hydrogen atom. The goal is to determine the wavelength of this emitted photon.

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Step-by-Step Solution:

#### 1. Understanding the Energy Levels in a Hydrogen Atom
The energy of an electron in the \( n \)-th energy level of a hydrogen atom is given by the formula:
\[
E_n = -\frac{R}{n^2}
\]
where:
- \( R \) is the Rydberg constant, approximately \( R = 2.18 \times 10^{-18} \, \text{J} \).
- \( n \) is the principal quantum number.

#### 2. Calculating the Energy Difference
When an electron transitions from the \( n = 4 \) level to the \( n = 2 \) level, the energy difference (\( \Delta E \)) between these two levels is:
\[
\Delta E = E_2 - E_4
\]

First, calculate \( E_2 \):
\[
E_2 = -\frac{R}{2^2} = -\frac{R}{4}
\]

Next, calculate \( E_4 \):
\[
E_4 = -\frac{R}{4^2} = -\frac{R}{16}
\]

Now, find the energy difference:
\[
\Delta E = E_2 - E_4 = \left( -\frac{R}{4} \right) - \left( -\frac{R}{16} \right)
\]
\[
\Delta E = -\frac{R}{4} + \frac{R}{16}
\]
\[
\Delta E = -\frac{4R}{16} + \frac{R}{16}
\]
\[
\Delta E = -\frac{3R}{16}
\]

Thus, the energy difference is:
\[
\Delta E = \frac{3R}{16}
\]

#### 3. Relating Energy to Wavelength
The energy of a photon is related to its wavelength (\( \lambda \)) by the equation:
\[
E = \frac{hc}{\lambda}
\]
where:
- \( h \) is Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{J·s} \).
- \( c \) is the speed of light, \( c = 3.00 \times 10^8 \, \text{m/s} \).

Rearranging for \( \lambda \):
\[
\lambda = \frac{hc}{E}
\]

Substitute \( E = \Delta E = \frac{3R}{16} \):
\[
\lambda = \frac{hc}{\frac{3R}{16}}
\]
\[
\lambda = \frac{16hc}{3R}
\]

#### 4. Substituting Known Values
- \( h = 6.626 \times 10^{-34} \, \text{J·s} \)
- \( c = 3.00 \times 10^8 \, \text{m/s} \)
- \( R = 2.18 \times 10^{-18} \, \text{J} \)

Calculate \( \lambda \):
\[
\lambda = \frac{16 \cdot (6.626 \times 10^{-34}) \cdot (3.00 \times 10^8)}{3 \cdot (2.18 \times 10^{-18})}
\]

First, compute the numerator:
\[
16 \cdot 6.626 \times 10^{-34} \cdot 3.00 \times 10^8 = 320.736 \times 10^{-26} = 3.20736 \times 10^{-24} \, \text{J·m}
\]

Next, compute the denominator:
\[
3 \cdot 2.18 \times 10^{-18} = 6.54 \times 10^{-18} \, \text{J}
\]

Now, divide the numerator by the denominator:
\[
\lambda = \frac{3.20736 \times 10^{-24}}{6.54 \times 10^{-18}}
\]
\[
\lambda = 0.490 \times 10^{-6} \, \text{m}
\]
\[
\lambda = 4.90 \times 10^{-7} \, \text{m}
\]

Convert to nanometers (\( 1 \, \text{nm} = 10^{-9} \, \text{m} \)):
\[
\lambda = 4.90 \times 10^{-7} \, \text{m} \times \frac{10^9 \, \text{nm}}{1 \, \text{m}} = 490 \, \text{nm}
\]

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Final Answer:

\[
\boxed{490 \, \text{nm}}
\]