Dual Nature of Matter

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149 of 200

Question 149 of 200

The light of two different frequencies whose photons have energies 3.8eV3.8eV and 1.4eV1.4eV respectively, illuminate a metallic surface whose work function is 0.6eV0.6eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

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Answer: (A) 2:12:1

Expression for maximum kinetic energy can be written as, (KE)max=E−ϕ{\left(KE\right)}_{max}=E-\phi
(KE)max,1=E1−ϕ=3.8−0.6=3.2eV{\left(KE\right)}_{max,1}={E}_{1}-\phi =3.8-0.6=3.2eV, (KE)max,2=E2−ϕ=1.4−0.6=0.8eV{\left(KE\right)}_{max,2}={E}_{2}-\phi =1.4-0.6=0.8eV
(KE)max,1(KE)max,2=3.20.8=4⇒4=12mv1212mv22\dfrac{{\left(KE\right)}_{max,1}}{{\left(KE\right)}_{max,2}}=\dfrac{3.2}{0.8}=4\Rightarrow 4=\dfrac{\dfrac{1}{2}{mv}_{1}^{2}}{\dfrac{1}{2}m{v}_{2}^{2}}
⇒v12v22=4\Rightarrow \dfrac{{v}_{1}^{2}}{{v}_{2}^{2}}=4 ⇒v1v2=2:1\Rightarrow \dfrac{{v}_{1}}{{v}_{2}}=2:1

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