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

1 : 5

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1 : 4

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1 : 2

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1 : 1

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Solution

The correct option is B $1 : 2$
Given : $ϕ = 0.5 e V$
Using $K . E = h ν - ϕ$ where $K . E = \frac{1}{2} m v^{2}$
For photon of energy, $h ν = 1 e V$
$K . E_{1} = 1 - 0.5 = 0.5 e V$
For photon of energy, $h ν = 2.5 e V$
$K . E_{2} = 2.5 - 0.5 = 2 e V$
Thus ratio of maximum speeds $\frac{v_{1}}{v_{2}} = \sqrt{\frac{K . E_{1}}{K . E_{2}}}$
$∴$ $\frac{v_{1}}{v_{2}} = \sqrt{\frac{0.5}{2}} = \frac{1}{2}$

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Light of two different frequencies whose protons 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 B 1:2Given : ϕ=0.5eVUsing K.E=hν−ϕ where K.E=12mv2For photon of energy, hν=1eVK.E1=1−0.5=0.5eV For photon of energy, hν=2.5eVK.E2=2.5−0.5=2eV Thus ratio of maximum speeds v1v2=√K.E1K.E2∴ v1v2=√0.52=12

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