The minimum energy required for the emission of photoelectron from the surface of a metal is `4.95 xx10^(-19)J` . Calculate the critical frequency of the photon required to eject the electron . ` h = 6.6 xx10^(-34) J` sec

Options

A. `7.5xx10^(14)s^(-1)`
B. `7.5xx10^(13)s^(-1)`
C. `7.5xx10^(16)s^(-1)`
D. `7.5xx10^(19)s^(-1)`

Answer

To solve the problem of calculating the critical frequency of the photon required to eject an electron from the surface of a metal, we can follow these steps:

### Step-by-step Solution

1. **Identify the given values**:
- Minimum energy required for the emission of a photoelectron, \( E = 4.95 \times 10^{-19} \, \text{J} \)
- Planck's constant, \( h = 6.6 \times 10^{-34} \, \text{J s} \)

2. **Use the formula relating energy and frequency**:
The energy of a photon is given by the equation:
\[ E = h \nu_0 \]
where \( \nu_0 \) is the critical frequency.

3. **Rearrange the formula to solve for the critical frequency**:
To find the critical frequency \( \nu_0 \), we rearrange the equation:
\[ \nu_0 = \frac{E}{h} \]

4. **Substitute the known values into the equation**:
\[ \nu_0 = \frac{4.95 \times 10^{-19} \, \text{J}}{6.6 \times 10^{-34} \, \text{J s}} \]

5. **Perform the calculation**:
\[ \nu_0 = \frac{4.95}{6.6} \times 10^{(-19 + 34)} \, \text{s}^{-1} \]
\[ \nu_0 = 0.75 \times 10^{15} \, \text{s}^{-1} \]
\[ \nu_0 = 7.5 \times 10^{14} \, \text{Hz} \]

6. **Final result**:
The critical frequency required to eject the electron is:
\[ \nu_0 = 7.5 \times 10^{14} \, \text{Hz} \]