Work funciton of a metal is 2.51eV. It threshold frequency is

Options

A. `5.9 xx 10^(14)` cycle/sec
B. `6.5 xx 10^(14)` cycle/sec
C. `9.4 xx 10^(14)` cycle/sec
D. `6.08 xx 10^(14)` cycle/sec

Answer

To find the threshold frequency of a metal given its work function, we can use the relationship between work function (Φ) and threshold frequency (ν₀). The work function is defined as the minimum energy required to remove an electron from the surface of a metal, and it is related to the threshold frequency by the equation:

\[ \Phi = h \cdot \nu_0 \]

Where:
- \(\Phi\) is the work function in joules,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(\nu_0\) is the threshold frequency in hertz (Hz).

### Step-by-Step Solution:

1. **Convert the Work Function from eV to Joules:**
The work function is given as \(2.51 \, \text{eV}\). To convert this to joules, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ \Phi = 2.51 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 4.016 \times 10^{-19} \, \text{J} \]

2. **Use the Work Function to Find the Threshold Frequency:**
Rearranging the equation \(\Phi = h \cdot \nu_0\) to solve for \(\nu_0\):
\[ \nu_0 = \frac{\Phi}{h} \]
Substituting the values we have:
\[ \nu_0 = \frac{4.016 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J s}} \]

3. **Calculate the Threshold Frequency:**
Performing the division:
\[ \nu_0 = 6.06 \times 10^{14} \, \text{Hz} \]

4. **Express the Frequency in Scientific Notation:**
The threshold frequency can also be expressed as:
\[ \nu_0 = 6.061 \times 10^{14} \, \text{Hz} \]

5. **Identify the Closest Option:**
If this is a multiple-choice question, you would select the option that corresponds to \(6.061 \times 10^{14} \, \text{Hz}\).

### Final Answer:
The threshold frequency of the metal is approximately \(6.061 \times 10^{14} \, \text{Hz}\).