When monochromatic light of wavelength 620 nm is used to illuminate a metallic surface, the maximum kinetic energy of photoelectrons emitted is 1 eV. Find the maximum kinetic energy of photoelectron emitted (in eV) if a wavelength of 155 nm is used on the same metallic surface. (hc = 1240 eV-nm)

Explanation

To solve the problem, we will use the principles of the photoelectric effect and Einstein's photoelectric equation, which states:

\[ E = \phi + KE_{\text{max}} \]

where:
- \( E \) is the energy of the incident photon,
- \( \phi \) is the work function of the metal,
- \( KE_{\text{max}} \) is the maximum kinetic energy of the emitted photoelectrons.

### Step 1: Calculate the energy of the incident photon for the first case.

Given:
- Wavelength \( \lambda_1 = 620 \, \text{nm} \)
- Maximum kinetic energy \( KE_{\text{max1}} = 1 \, \text{eV} \)

Using the formula for energy of a photon:
\[ E = \frac{hc}{\lambda} \]

Substituting the values:
\[ E_1 = \frac{1240 \, \text{eV-nm}}{620 \, \text{nm}} = 2 \, \text{eV} \]

### Step 2: Find the work function \( \phi \).

From the photoelectric equation:
\[ E_1 = \phi + KE_{\text{max1}} \]

Substituting the known values:
\[ 2 \, \text{eV} = \phi + 1 \, \text{eV} \]

Solving for \( \phi \):
\[ \phi = 2 \, \text{eV} - 1 \, \text{eV} = 1 \, \text{eV} \]

### Step 3: Calculate the energy of the incident photon for the second case.

Given:
- Wavelength \( \lambda_2 = 155 \, \text{nm} \)

Using the same formula for energy of a photon:
\[ E_2 = \frac{hc}{\lambda_2} \]

Substituting the values:
\[ E_2 = \frac{1240 \, \text{eV-nm}}{155 \, \text{nm}} \approx 8 \, \text{eV} \]

### Step 4: Find the maximum kinetic energy \( KE_{\text{max2}} \) for the second case.

Using the photoelectric equation again:
\[ E_2 = \phi + KE_{\text{max2}} \]

Substituting the known values:
\[ 8 \, \text{eV} = 1 \, \text{eV} + KE_{\text{max2}} \]

Solving for \( KE_{\text{max2}} \):
\[ KE_{\text{max2}} = 8 \, \text{eV} - 1 \, \text{eV} = 7 \, \text{eV} \]

### Final Answer:
The maximum kinetic energy of the photoelectron emitted when a wavelength of 155 nm is used is **7 eV**.

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