The difference between threshold wavelengths for two metal surfaces A and B having work function \(Φ_A\)= 9 eV and \(Φ_B\) = 4.5 eV in nm is:
{Given, hc = 1242 eV nm}
{Given, hc = 1242 eV nm}
Show Hint
{Threshold Wavelength}: The threshold wavelength is the longest wavelength of light that can eject electrons from a metal surface. It is inversely proportional to the work function of the metal. A lower work function results in a longer threshold wavelength.
- 264
- 540
- 276
- 138
The Correct Option is D
Solution and Explanation
The threshold wavelength (\( \lambda_{\text{threshold}} \)) for a metal surface is related to its work function (\( \phi \)) by the equation:
\[ \lambda_{\text{threshold}} = \frac{hc}{\phi} \]where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \phi \) is the work function of the metal.
Given:
\[ hc = 1242 \, \text{eV nm} \] \[ \phi_A = 9 \, \text{eV} \] \[ \phi_B = 4.5 \, \text{eV} \]Step 1: Calculate Threshold Wavelengths for Both Metals
For Metal A:
\[ \lambda_{\text{threshold}, A} = \frac{1242 \, \text{eV nm}}{9 \, \text{eV}} = 138 \, \text{nm} \]For Metal B:
\[ \lambda_{\text{threshold}, B} = \frac{1242 \, \text{eV nm}}{4.5 \, \text{eV}} = 276 \, \text{nm} \]Step 2: Determine the Difference Between Threshold Wavelengths
\[ \Delta \lambda = \lambda_{\text{threshold}, B} - \lambda_{\text{threshold}, A} = 276 \, \text{nm} - 138 \, \text{nm} = 138 \, \text{nm} \]Therefore, the difference between the threshold wavelengths for metal surfaces A and B is 138 nm, which corresponds to option (4).
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