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Q.

Photons of energies 1 eV and 2 eV are successively incident on a metallic surface of work function 0·5 eV. The ratio of kinetic energy of most energetic photoelectrons in the two cases will be

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a

1 : 2

b

1 : 1

c

1 : 3

d

1 : 4

answer is C.

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Detailed Solution

Kinetic energy due to 1eV photon will be, K1= 1 - 0.5 (the energy used to overcome the work function) = 0.5
Kinetic energy due to 2 eV photon will be, K2 = 2 - 0.5 = 1.5 
Therefore. K1/K2 = 0.5/1.5 = 1/3

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