The difference between threshold wavelengths for two metal surfaces A and B having work function φA = 9 eV and φB = 4.5 eV in nm is: Given, hc = 1242 eVnm (a) 264 (b) 138 (c) 540 (d) 276

Answer: (b). λ₀ = (hc)/(φ) λA = 1242/9 = 138 nm, λB = (1242)/(4.5) = 276 nm Difference = 276 - 138 = 138 nm So the answer is (b) 138.

The difference between threshold wavelengths for two metal surfaces A… | JEE Main 2023 (13 Apr, Shift 1)

Q72JEE Main 2023 · 13 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation

The difference between threshold wavelengths for two metal surfaces $\text{A}$ and $\text{B}$ having work function $\phi_A = 9\text{ eV}$ and $\phi_B = 4.5\text{ eV}$ in $\text{nm}$ is:

$\{$ Given, hc $= 1242\text{ eVnm}\}$

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Answer: (b)

$\lambda_0 = \dfrac{hc}{\phi}$
$\lambda_A = \dfrac{1242}{9} = 138\text{ nm}$ , $\lambda_B = \dfrac{1242}{4.5} = 276\text{ nm}$
Difference $= 276 - 138 = 138\text{ nm}$
So the answer is (b) 138.

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