Question

The difference between threshold wavelengths for two metal surfaces \( A \) and \( B \) having work functions \( \phi_A = 9 \, \text{eV} \) and \( \phi_B = 4.5 \, \text{eV} \) is: \[ \text{(Given: } h c = 1242 \, \text{eV nm)} \]

• A. \( 264 \, \text{nm} \)
• B. \( 138 \, \text{nm} \)
• C. \( 276 \, \text{nm} \)
• D. \( 540 \, \text{nm} \)

Hint:

The threshold wavelength decreases as the work function increases. Use the formula \( \lambda = \frac{1242}{\phi} \) to find the relationship and ensure all units are consistent.

Answer 1

The Correct Option is B

The wavelength (\( \lambda \)) associated with the threshold energy is related to the work function (\( \phi \)) as: \[ \lambda = \frac{hc}{\phi}, \] where \( h \) is Planck’s constant, \( c \) is the speed of light, and \( \phi \) is the work function in eV. 

Step 1: Calculate Wavelengths For \( \phi_A = 9 \, \text{eV} \): \[ \lambda_A = \frac{1242}{9} = 138 \, \text{nm}. \] For \( \phi_B = 4.5 \, \text{eV} \): \[ \lambda_B = \frac{1242}{4.5} = 276 \, \text{nm}. \] 

Step 2: Calculate the Difference in Wavelengths \[ \Delta \lambda = \lambda_B - \lambda_A = 276 - 138 = 138 \, \text{nm}. \] 

Final Answer: \[ \boxed{138 \, \text{nm}} \]