A hydrogen atom in ground state obsebe a photon of ultraviolet radition of wavelength `50 nm` Assuming that the entire photon energy is taken up by the electron with what kinetic energy will the up by the electron with what kinetic energy will the electron be ejected?

Explanation

To solve the problem of determining the kinetic energy of an electron ejected from a hydrogen atom after absorbing a photon of ultraviolet radiation with a wavelength of 50 nm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Wavelength of the photon, \( \lambda = 50 \, \text{nm} = 50 \times 10^{-9} \, \text{m} \)

2. **Calculate the Energy of the Photon**:
The energy \( E \) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) (Planck's constant) is approximately \( 6.626 \times 10^{-34} \, \text{Js} \) and \( c \) (speed of light) is approximately \( 3 \times 10^8 \, \text{m/s} \).

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{50 \times 10^{-9} \, \text{m}} \]

\[ E = \frac{1.9878 \times 10^{-25}}{50 \times 10^{-9}} = 3.9756 \times 10^{-18} \, \text{J} \]

To convert this energy into electron volts (eV), we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{3.9756 \times 10^{-18} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 24.85 \, \text{eV} \]

3. **Determine the Work Function**:
The work function \( \phi \) for a hydrogen atom in the ground state (n=1) is given by:
\[ \phi = 13.6 \, \text{eV} \]

4. **Calculate the Kinetic Energy of the Ejected Electron**:
The kinetic energy \( KE \) of the ejected electron can be calculated using the equation:
\[ KE = E - \phi \]
Substituting the values:
\[ KE = 24.85 \, \text{eV} - 13.6 \, \text{eV} = 11.25 \, \text{eV} \]

5. **Final Answer**:
The kinetic energy with which the electron will be ejected is approximately \( 11.25 \, \text{eV} \).