Question

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
(A) 1:2
(B) 1:1
(C) 1:3
(D) 1:4​

Answer 1

Answer:

option (c) is the answer

Answer 2

C) 1:3

Explanation:

Given:

energy of first photon, $E_1=1eV$

energy of second photon, $E_2=2\ eV$

work function of the surface, $W=0.5\ eV$

We know that:

$E=W+KE$

where:

$E=$ energy of photon

$KE=$ kinetic energy of the photon $=\frac{1}{2}m.v^2$

m= mass of photon

v= velocity of photon

Now for first photon:

$E_1=W+KE_1$

$1=0.5+KE_1$

$KE_1=0.5\ eV$

Now for second photon:

$E_2=W+KE_2$

$2=0.5+KE_2$

$KE_2=1.5\ eV$

Now the ratio:

$\frac{KE_1}{KE_2} =\frac{0.5}{1.5} =\frac{1}{3}$

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TOPIC: photons

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