The radiation emitted by an electronic transition from n=3 to n=2 in a hydrogen atom is used to induce photoelectric effect on a metal surface. Find the work function (in eV0 of the surface, if the maximum kinetic energy of ejected electrons is measured tobe `1.376xx10^(-19)J`

Explanation

To solve the problem step by step, we will follow the principles of quantum mechanics and the photoelectric effect.

### Step 1: Calculate the energy of the photon emitted during the electronic transition from n=3 to n=2.

The energy of a photon emitted during a transition in a hydrogen atom can be calculated using the formula:

\[ E = -13.6 \, \text{eV} \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]

where \( n_1 = 2 \) and \( n_2 = 3 \).

Substituting the values:

\[ E = -13.6 \, \text{eV} \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating the fractions:

\[ E = -13.6 \, \text{eV} \left( \frac{1}{4} - \frac{1}{9} \right) \]

Finding a common denominator (36):

\[ E = -13.6 \, \text{eV} \left( \frac{9 - 4}{36} \right) = -13.6 \, \text{eV} \left( \frac{5}{36} \right) \]

Calculating the energy:

\[ E = -13.6 \times \frac{5}{36} \approx -1.89 \, \text{eV} \]

### Step 2: Convert the maximum kinetic energy from Joules to electron volts.

The maximum kinetic energy of the ejected electrons is given as:

\[ KE = 1.376 \times 10^{-19} \, \text{J} \]

To convert Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\[ KE = \frac{1.376 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 0.86 \, \text{eV} \]

### Step 3: Calculate the work function of the metal surface.

The work function \( \phi \) can be calculated using the equation:

\[ \phi = E - KE \]

Substituting the values we have:

\[ \phi = 1.89 \, \text{eV} - 0.86 \, \text{eV} = 1.03 \, \text{eV} \]

### Conclusion

The work function of the metal surface is approximately:

\[ \phi \approx 1.03 \, \text{eV} \]