Wavelengths belonging to Balmer series lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work function is 2.0 eV. Find ( in eV ) the maximum kinetic energy of the emitted photoelectrons. `("Take hc = 1242 eV nm.")`

Answer

To solve the problem, we need to find the maximum kinetic energy of the emitted photoelectrons when light of specific wavelengths is incident on a metal surface. The work function of the metal is given, and we will use the relationship between energy, wavelength, and kinetic energy.

### Step-by-Step Solution:

1. **Identify the Wavelength Range**:
The problem states that the wavelengths belonging to the Balmer series lie between 450 nm and 750 nm. To find the maximum kinetic energy, we need to use the minimum wavelength (450 nm) because energy is inversely proportional to wavelength.

2. **Calculate the Energy of the Incident Light**:
The energy \(E\) of the incident light can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant and \(c\) is the speed of light. Given that \(hc = 1242 \, \text{eV nm}\), we can substitute \(h\) and \(c\) into the equation:
\[ E = \frac{1242 \, \text{eV nm}}{450 \, \text{nm}} \]

3. **Perform the Calculation**:
\[ E = \frac{1242}{450} \approx 2.76 \, \text{eV} \]

4. **Determine the Maximum Kinetic Energy**:
The maximum kinetic energy \(K_{\text{max}}\) of the emitted photoelectrons can be found using the equation:
\[ K_{\text{max}} = E - \phi \]
where \(\phi\) is the work function of the metal. Given that the work function is \(2.0 \, \text{eV}\):
\[ K_{\text{max}} = 2.76 \, \text{eV} - 2.0 \, \text{eV} \]

5. **Final Calculation**:
\[ K_{\text{max}} = 0.76 \, \text{eV} \]

### Conclusion:
The maximum kinetic energy of the emitted photoelectrons is \(0.76 \, \text{eV}\).

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