The work function of a metal is `4.2 eV` , its threshold wavelength will be

Options

A. `4000 Å`
B. `3500 Å`
C. `2955 Å`
D. `2500 Å`

Explanation

To find the threshold wavelength of a metal given its work function, we can use the relationship between energy and wavelength. The work function (Φ) is the minimum energy required to remove an electron from the surface of a metal, and it is related to the threshold wavelength (λ₀) by the equation:

\[ E = \frac{hc}{\lambda_0} \]

Where:
- \( E \) is the energy (in electron volts)
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
- \( \lambda_0 \) is the threshold wavelength (in meters)

However, for convenience, we can use the simplified formula that relates the work function in electron volts to the wavelength in angstroms:

\[ \lambda_0 = \frac{12375}{\Phi} \]

Where:
- \( \lambda_0 \) is in angstroms
- \( \Phi \) is the work function in electron volts

### Step-by-Step Solution:

1. **Identify the Work Function**: The work function of the metal is given as \( \Phi = 4.2 \, \text{eV} \).

2. **Use the Formula for Threshold Wavelength**: We will use the formula \( \lambda_0 = \frac{12375}{\Phi} \).

3. **Substitute the Work Function into the Formula**:
\[ \lambda_0 = \frac{12375}{4.2} \]

4. **Calculate the Wavelength**:
\[ \lambda_0 = \frac{12375}{4.2} \approx 2957.14 \, \text{Å} \]

5. **Round the Result**: The threshold wavelength can be approximated to \( \lambda_0 \approx 2955 \, \text{Å} \).

### Final Answer:
The threshold wavelength of the metal is approximately **2955 Å**.

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