The difference between threshold wavelengths for two metal surface A and B haveing work function` phi _A=9eV and phi _b =4*5 eV` in nm is
{Given hc= 1242 eV nm }.

Options

A. `264`
B. `540`
C. `138`
D. `276`

Answer

To find the difference between the threshold wavelengths for two metal surfaces A and B with given work functions, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between work function and threshold wavelength**:
The threshold wavelength (\( \lambda_{th} \)) is related to the work function (\( \phi \)) by the equation:
\[ \phi = \frac{hc}{\lambda_{th}} \]
Rearranging this gives:
\[ \lambda_{th} = \frac{hc}{\phi} \]

2. **Calculate the threshold wavelength for metal A**:
Given the work function for metal A is \( \phi_A = 9 \, \text{eV} \) and using \( hc = 1242 \, \text{eV nm} \):
\[ \lambda_{th_A} = \frac{1242 \, \text{eV nm}}{9 \, \text{eV}} = \frac{1242}{9} \, \text{nm} \]
\[ \lambda_{th_A} = 138.0 \, \text{nm} \]

3. **Calculate the threshold wavelength for metal B**:
Given the work function for metal B is \( \phi_B = 4.5 \, \text{eV} \):
\[ \lambda_{th_B} = \frac{1242 \, \text{eV nm}}{4.5 \, \text{eV}} = \frac{1242}{4.5} \, \text{nm} \]
\[ \lambda_{th_B} = 276.0 \, \text{nm} \]

4. **Find the difference between the threshold wavelengths**:
Now, we can find the difference:
\[ \Delta \lambda_{th} = \lambda_{th_B} - \lambda_{th_A} = 276.0 \, \text{nm} - 138.0 \, \text{nm} \]
\[ \Delta \lambda_{th} = 138.0 \, \text{nm} \]

5. **Final Result**:
The difference between the threshold wavelengths for the two metal surfaces A and B is:
\[ \Delta \lambda_{th} = 138.0 \, \text{nm} \]