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

1:2

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1:1

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1:5

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1:4

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Solution

The correct option is A 1:2
$E_{1} = 1 e V ϕ_{1} = ϕ_{2} = 0.5 e V$
$E_{2} = 2.5 e V$
Equation of photoelectric effect
$E_{1} = ϕ_{1} + K E_{1}, E_{2} = ϕ_{2} + K E_{2}$
$1 = 0.5 + \frac{1}{2} m v_{1} 2, 2.5 = 0.5 + \frac{1}{2} m v_{2}^{2}$
$\frac{0.5 \times 2}{m} = v_{1}^{2}, \frac{2 \times 2}{m} = v_{2}^{2}$
${[\frac{v_{1}}{v_{2}}]}^{2} = \frac{1}{4}$
$\frac{v_{1}}{v_{2}} = \frac{1}{2}$

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

The correct option is A 1:2E1=1eV ϕ1=ϕ2=0.5eV E2=2.5eV Equation of photoelectric effect E1=ϕ1+KE1,E2=ϕ2+KE2 1=0.5+12mv12,2.5=0.5+12mv22 0.5×2m=v21,2×2m=v22 [v1v2]2=14 v1v2=12