Question

Hi !

Solve this question JEE mains 2018 .

Surface of certain metal illuminated with light of wavelength L1 = 350nm and then by light of wavelength ,L2 = 540nm . if Is found that the maximum speed of the photo Electrons in the two cases differ by a factor of 2.The work function of the metal (in ev) I'd close to ? ​

Answer 1

Answer:

1.88

Explanation :-

Given that surface of metal is first illuminated with light of wavelength L1 = 350nm and then L 2 = 540nm . it is found that the maximum speed of photo Electrons In the two cases differ by 2

so first of all let the L1 wavelength' light has speed V

And , E1 = 1240/350ev

E2'= 1240/540ev

so that , L2 wavelength's light will be 2v

Now, according to Einstein photoelectric equation .

E - ¢ = 1/2Mv²

E1- ¢ = 1/mv² _________(1)

E2 - ¢ = 1/2m(2v)² _________(2)

Dividing (1) to (2)

We get,

E1-¢ /E2-¢ = 4

E1-¢ = 4E2 - 4¢

¢ = 4E2 - E1/3

= 1240(4/540-1/350) = 1.88 ev Answer

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Hope it's helpful

Answer 2

Answer:

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