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Q.

K₁ and K₂ are the maximum kinetic energies of the photo electrons emitted when light of wave lengths λ₁ and λ₂ respectively are incident on a metallic surface. If ${\lambda _1} = 3{\lambda _2}\$ then

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a

${K_1} < \frac{{{K_2}}}{3}\$

b

K₂ = 3K₁

c

${K_1} > \frac{{{K_2}}}{3}\$

d

K₁ > 3K₂

answer is B.

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Detailed Solution

$K{E_{\max }} = E - {W_0}\,;$

$\,\therefore {K_1} = \frac{{hc}}{\lambda_1 } - {W_o} \to (1)\$

$and\,\,{K_2} = \frac{{hc}}{{{\lambda _2}}} - {W_o}\,;$ $\therefore {K_2} = \frac{{hc}}{{\left(\frac{\lambda _1}{3}\right)}} - {W_o}\$

$\therefore$ K₂ = 3[K₁ + W_o] – W_o ;

K₂ = 3K₁ + 2W_o

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