The threshold wavelength of a metal having work function 2 eV is 6000 Å . What will be the threshold wavelength for the metal having a work function of 6 eV ?

Options

A. 3000 Å
B. 2000 Å
C. 8000 Å
D. 18000 Å

Explanation

To solve the problem, we will use the relationship between the work function (W) and the threshold wavelength (λ) of a metal. The work function is related to the threshold wavelength by the equation:

\[ W = \frac{hc}{\lambda} \]

Where:
- \( W \) is the work function in electron volts (eV),
- \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \) eV·s),
- \( c \) is the speed of light (\( 3 \times 10^8 \) m/s),
- \( \lambda \) is the threshold wavelength in meters.

### Step-by-Step Solution:

1. **Identify the given values:**
- For the first metal:
- Work function \( W_1 = 2 \) eV
- Threshold wavelength \( \lambda_1 = 6000 \) Å (which is \( 6000 \times 10^{-10} \) m)
- For the second metal:
- Work function \( W_2 = 6 \) eV
- Threshold wavelength \( \lambda_2 \) is unknown.

2. **Use the relationship between work function and threshold wavelength:**
From the equation \( W = \frac{hc}{\lambda} \), we can rearrange it to find the threshold wavelength:
\[ \lambda = \frac{hc}{W} \]

3. **Set up the ratio of the two metals:**
Since the work function is inversely proportional to the wavelength, we can write:
\[ \frac{W_1}{W_2} = \frac{\lambda_2}{\lambda_1} \]

4. **Substitute the known values into the equation:**
\[ \frac{2 \text{ eV}}{6 \text{ eV}} = \frac{\lambda_2}{6000 \text{ Å}} \]

5. **Cross-multiply to solve for \( \lambda_2 \):**
\[ 2 \times 6000 = 6 \times \lambda_2 \]
\[ 12000 = 6 \lambda_2 \]
\[ \lambda_2 = \frac{12000}{6} = 2000 \text{ Å} \]

6. **Conclusion:**
The threshold wavelength for the metal with a work function of 6 eV is \( \lambda_2 = 2000 \) Å.

### Final Answer:
The threshold wavelength for the metal having a work function of 6 eV is **2000 Å**.