Problem

Find the longest wavelength of light capable of ionizing a hydrogen atom and calculate the energy needed to ionize a hydrogen atom.

Data

• Initial Energy Level ($n$): $\infty$
• Final Energy Level ($p$): $1$
• 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 ($E$)

Prerequisite Concepts

1. Wavelength Formula (Hydrogen Atom):
   The wavelength for ionization can be calculated using:

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

Where:

• $R_H$: Rydberg constant
• $p$: Lower energy level
• $n$: Higher energy level ($n > p$)

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

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

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

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

Solution

Part 1: Calculate the Wavelength ($\lambda$)

Using the formula for wavelength:

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

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

$$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$$

Since $\frac{1}{\infty^2} = 0$:

$$\frac{1}{\lambda} = 1.0974 \times 10^{7}$$ $$\lambda = \frac{1}{1.0974 \times 10^{7}} \, \mathrm{m}$$ $$\lambda = 9.11244 \times 10^{-8} \, \mathrm{m}$$

Convert to nanometers:

$$\lambda = 91.1 \, \mathrm{nm}$$

Part 2: 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 $\lambda = 9.11244 \times 10^{-8} \, \mathrm{m}$:

$$E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^{8}}{9.11244 \times 10^{-8}}$$ $$E = 2.1819 \times 10^{-18} \, \mathrm{J}$$

Convert to electronvolts:

$$E = \frac{2.1819 \times 10^{-18}}{1.602 \times 10^{-19}}$$ $$E = 13.619 \, \mathrm{eV}$$

Answer

1. Longest Wavelength:

$$\lambda = 91.1 \, \mathrm{nm}$$

2. Energy Needed to Ionize:

$$E = 13.619 \, \mathrm{eV}$$

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