The threshold wavelength for a metal is 5000 Å . The maximum K.E. of the photoelectrons emitted when ultraviolet ligh of wavelength 2500 Å falls on it is

Options

A. 8.28 eV
B. 4.28 e V
C. 2.48 eV
D. 24. 8 e V

Answer

To solve the problem, we need to calculate the maximum kinetic energy (KE) of the photoelectrons emitted when ultraviolet light of wavelength 2500 Å falls on a metal with a threshold wavelength of 5000 Å.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Threshold wavelength (\( \lambda_0 \)) = 5000 Å = \( 5000 \times 10^{-10} \) m = \( 5.0 \times 10^{-7} \) m
- Wavelength of incident light (\( \lambda \)) = 2500 Å = \( 2500 \times 10^{-10} \) m = \( 2.5 \times 10^{-7} \) m

2. **Calculate the Energy of the Incident Photon:**
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \) J·s
- \( c \) (speed of light) = \( 3.0 \times 10^8 \) m/s

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \text{ J·s})(3.0 \times 10^8 \text{ m/s})}{2.5 \times 10^{-7} \text{ m}} \]

3. **Calculate the Work Function Energy:**
Similarly, the work function energy (\( \phi \)) can be calculated using the threshold wavelength:
\[ \phi = \frac{hc}{\lambda_0} \]
Substituting the values:
\[ \phi = \frac{(6.626 \times 10^{-34} \text{ J·s})(3.0 \times 10^8 \text{ m/s})}{5.0 \times 10^{-7} \text{ m}} \]

4. **Calculate Maximum Kinetic Energy:**
The maximum kinetic energy of the emitted photoelectrons is given by:
\[ KE_{\text{max}} = E - \phi \]
Substituting the values calculated in steps 2 and 3.

5. **Convert the Kinetic Energy to Electron Volts:**
To convert the energy from joules to electron volts, use the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Thus, to convert:
\[ KE_{\text{max (eV)}} = \frac{KE_{\text{max (J)}}}{1.6 \times 10^{-19}} \]

6. **Final Calculation:**
After performing the calculations, we find that:
\[ KE_{\text{max}} \approx 2.48 \text{ eV} \]

### Final Answer:
The maximum kinetic energy of the photoelectrons emitted is approximately **2.48 eV**.

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