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

1:2

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1:5

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1:1

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1:4

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Solution

The correct option is A
1:2

$E = ϕ_{0} + K_{m a x} 1 = 0.5 + K_{1} = K_{1} \Rightarrow 0.5 2.5 = 0.5 + K_{2} \Rightarrow K_{2} = 2 \frac{K_{1}}{K_{2}} = \frac{0.5}{2} = \frac{1}{4} (\frac{v_{1}}{v_{2}}) = \sqrt{\frac{K_{1}}{K_{2}}} = \frac{1}{2}$

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

The correct option is A 1:2 E=ϕ0+Kmax1=0.5+K1=K1⇒0.52.5=0.5+K2⇒K2=2K1K2=0.52=14(v1v2)=√K1K2=12

The correct option is A
1:2

$E = ϕ_{0} + K_{m a x} 1 = 0.5 + K_{1} = K_{1} \Rightarrow 0.5 2.5 = 0.5 + K_{2} \Rightarrow K_{2} = 2 \frac{K_{1}}{K_{2}} = \frac{0.5}{2} = \frac{1}{4} (\frac{v_{1}}{v_{2}}) = \sqrt{\frac{K_{1}}{K_{2}}} = \frac{1}{2}$