What will be the minimum frequency of light source to get photocurrent, from a metal surface having work function 2 eV ?

Options

A. `4.8xx10^(14)Hz`
B. `5.8xx10^(14)Hz`
C. `7.8xx10^(14)Hz`
D. `2.8xx10^(14)Hz`

Answer

To find the minimum frequency of light required to produce photocurrent from a metal surface with a work function of 2 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Work Function**:
The work function (ϕ) is the minimum energy required to remove an electron from the surface of a metal. In this case, the work function is given as 2 eV.

2. **Convert Work Function to Joules**:
Since the energy is given in electron volts (eV), we need to convert it to joules (J).
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Therefore,
\[ \phi = 2 \text{ eV} = 2 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-19} \text{ J} \]

3. **Use the Photoelectric Equation**:
The photoelectric effect can be described by the equation:
\[ K.E. = h\nu - \phi \]
where \( K.E. \) is the kinetic energy of the emitted electron, \( h \) is Planck's constant, and \( \nu \) is the frequency of the incident light. For photocurrent to be generated, the kinetic energy must be greater than or equal to zero:
\[ 0 \leq h\nu - \phi \]
Rearranging gives:
\[ h\nu \geq \phi \]

4. **Substitute Known Values**:
Planck's constant \( h \) is approximately \( 6.63 \times 10^{-34} \text{ J s} \). We can substitute the values into the equation:
\[ \nu \geq \frac{\phi}{h} = \frac{3.2 \times 10^{-19} \text{ J}}{6.63 \times 10^{-34} \text{ J s}} \]

5. **Calculate the Frequency**:
Performing the division:
\[ \nu \geq \frac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 4.82 \times 10^{14} \text{ Hz} \]

6. **Conclusion**:
The minimum frequency of the light source required to produce photocurrent from the metal surface is approximately \( 4.82 \times 10^{14} \text{ Hz} \).

### Final Answer:
The minimum frequency of the light source to get photocurrent is \( 4.82 \times 10^{14} \text{ Hz} \).

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