If the photoelectric work function of a metal is 5 eV , the threshold frequency of the metal is

Options

A. `2.2 xx 10^(15)` Hz
B. `1.2 xx 10^(15)` Hz
C. `1.5 xx 10^(15)` Hz
D. `5.1 xx 10^(15)` Hz

Answer

To find the threshold frequency of a metal given its photoelectric work function, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Work Function**: The work function (Φ) is the minimum energy required to eject an electron from the surface of a metal. In this case, the work function is given as 5 eV.

2. **Convert Work Function to Joules**: Since the work function is given in electron volts (eV), we need to convert it to joules (J). The conversion factor is:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Therefore,
\[ \Phi = 5 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 8 \times 10^{-19} \text{ J} \]

3. **Use the Relation Between Work Function and Threshold Frequency**: The relationship between the work function and the threshold frequency (ν₀) is given by the equation:
\[ \Phi = h \nu_0 \]
where \(h\) is Planck's constant, approximately \(6.63 \times 10^{-34} \text{ J s}\).

4. **Rearrange the Equation to Solve for Threshold Frequency**: To find the threshold frequency, rearrange the equation:
\[ \nu_0 = \frac{\Phi}{h} \]

5. **Substitute the Values**: Now substitute the values we have:
\[ \nu_0 = \frac{8 \times 10^{-19} \text{ J}}{6.63 \times 10^{-34} \text{ J s}} \]

6. **Calculate the Threshold Frequency**: Performing the division:
\[ \nu_0 \approx 1.21 \times 10^{15} \text{ Hz} \]

### Final Answer
The threshold frequency of the metal is approximately \(1.21 \times 10^{15} \text{ Hz}\).

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