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 find the critical frequency of the photon required to eject an electron from the surface of a metal, we can use the relationship between energy, frequency, and Planck's constant. The energy required to emit a photoelectron is given as \( E = 4.95 \times 10^{-19} \, \text{J} \), and Planck's constant \( h \) is given as \( 6.625 \times 10^{-34} \, \text{J s} \).

### Step-by-Step Solution:

1. **Understand the relationship**: The energy of a photon can be expressed as:
\[ E = h \nu \]
where \( E \) is the energy of the photon, \( h \) is Planck's constant, and \( \nu \) is the frequency of the photon.

2. **Rearrange the formula**: To find the frequency \( \nu \), we can rearrange the equation:
\[ \nu = \frac{E}{h} \]

3. **Substitute the values**: Now, substitute the given values into the equation:
\[ \nu = \frac{4.95 \times 10^{-19} \, \text{J}}{6.625 \times 10^{-34} \, \text{J s}} \]

4. **Calculate the frequency**: Perform the division:
\[ \nu = \frac{4.95 \times 10^{-19}}{6.625 \times 10^{-34}} \approx 7.47 \times 10^{14} \, \text{Hz} \]

5. **Round the answer**: Rounding to two decimal places, we get:
\[ \nu \approx 7.5 \times 10^{14} \, \text{Hz} \]

### Final Answer:
The critical frequency of the photon required to eject the electron is approximately \( 7.5 \times 10^{14} \, \text{Hz} \).