# When Photons Of Energy 4.25Ev Strike The Surface Of A Metal A, The Ejected Photoelectrons Have Maximum Kinetic Energy Of K Aev And De Broglie Wavelength λA. The Maximum Kinetic Energy Of Photo Electrons Liberated From Another Metal B By Photons Of Energy 4.70Ev Is K B =(Ka −1.5)Ev. If The De Broglie Wavelength Of These Photoelectrons Is λ B=2λ A Then

De Broglie wavelength $$\lambda = \frac{h}{p}$$

Kinetic energy K = $$\frac{p^{2}}{2m}$$

Therefore,

$$\Rightarrow \frac{K_{A}}{K_{B}} = \frac{P_{A}^{2}}{P_{B}^{2}}$$ $$\Rightarrow \frac{\lambda_{B}^{2}}{\lambda_{A}^{2}} = 4$$

As $$\lambda_{B} = 2 \lambda_{A}$$

Given $$K_{B} = (K_{A} -1.5)$$ $$\Rightarrow 4K_{B} – 1.5$$ $$\Rightarrow K_{B} = 0.5eV$$

Thus, $$K_{A} = 4 * 0.5 = 2.00eV$$

From photo-electric effect:

Work function of A = 4.25 -$$(K_{A}$$ $$\Rightarrow 4.25 – 2.00 = 2.25eV$$

Work function of B = 4.70 – $$(K_{B}$$ $$\Rightarrow 4.70 – 0.5 = 4.20eV$$

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