Light of two different frequencies whose photons have energies $1 eV and 2.5 eV$ respectively illuminate a metallic surface whose work function is $0.5 eV$ successively. Ratio of maximum speeds emitted electrons will be :

1 : 1

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1 : 4

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1 : 2

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1 : 5

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Solution

The correct option is C $1 : 2$
Given:
$E_{1} = 1 eV, E_{2} = 2.5 eV$
work function $= 0.5 eV$
We know that,
$K . E = ϕ - ϕ_{0}$
$K . E_{1} = 1 eV - 0.5 eV = 0.5 eV$
$K . E_{2} = 2.5 eV - 0.5 eV = 2 eV$
$\frac{K . E_{1}}{K . E_{2}} = \frac{0.5 eV}{2 eV} = \frac{1}{4};$
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$ .
Hence, option $(B)$ is correct.

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Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds emitted electrons will be :

The correct option is C 1:2Given: E1=1 eV , E2=2.5 eV work function =0.5 eV We know that, K.E=ϕ−ϕ0 K.E1=1 eV−0.5 eV=0.5 eV K.E2=2.5 eV−0.5 eV=2 eV K.E1K.E2=0.5 eV2 eV=14; v1v2=√14=12. Hence, option (B) is correct.

Light of two different frequencies whose photons have energies $1 eV and 2.5 eV$ respectively illuminate a metallic surface whose work function is $0.5 eV$ successively. Ratio of maximum speeds emitted electrons will be :