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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