Question

The maximum wavelength of incident radiation required to ionize a hydrogen atom in its ground state is nearly

• A. 912 nm
• B. 1215 $\text{\AA}$
• C. 912 $\text{\AA}$
• D. 1215 nm

Hint:

The ionization energy of hydrogen in the ground state is $13.6 \text{ eV}$. The product of Planck's constant and the speed of light is a useful constant for quickly calculating wavelength from energy: $hc \approx 1240 \text{ eV}\cdot\text{nm}$ or $12400 \text{ eV}\cdot\text{\AA}$.

Answer 1

The Correct Option is C

Step 1: Relate ionization to the energy of the incident radiation.
Ionization of a hydrogen atom in its ground state ($n=1$) means exciting the electron from $n=1$ to $n=\infty$ (free electron). The minimum energy required for this process is the ionization energy ($E_{ion}$). The energy of the incident radiation must be equal to or greater than $E_{ion}$.

Step 2: Find the ionization energy for the hydrogen atom.
The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is $E_n = -\frac{13.6}{n^2} \text{ eV}$. The ground state energy ($n=1$) is $E_1 = -13.6 \text{ eV}$. The energy of the ionized state ($n=\infty$) is $E_\infty = 0 \text{ eV}$. \[ E_{ion} = E_\infty - E_1 = 0 - (-13.6 \text{ eV}) = 13.6 \text{ eV}. \]

Step 3: Relate the energy to the maximum wavelength.
The energy of a photon is $E = \frac{hc}{\lambda}$. For the maximum wavelength ($\lambda_{max}$), the energy must be minimum and equal to $E_{ion}$. \[ \lambda_{max} = \frac{hc}{E_{ion}}. \]

Step 4: Use the shortcut formula for $\lambda_{max$.}
The product $hc$ is often given in electron-volt meters: $hc \approx 1240 \text{ eV}\cdot\text{nm}$ or $12400 \text{ eV}\cdot\text{\AA}$. We use the $\text{\AA}$ version since the options are in $\text{\AA}$. \[ \lambda_{max} = \frac{12400 \text{ eV}\cdot\text{\AA}}{13.6 \text{ eV}}. \] \[ \lambda_{max} \approx 911.76 \text{ \AA}. \]

Step 5: Conclude the result.
The maximum wavelength is nearly $912 \text{ \AA}$.