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JEE Main 202117 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
17 March
Shift
Evening Shift
Subject
Physics
Chapter
Dual Nature of Matter

JEE Main 2021 · 17 March · Evening Shift · Physics · Dual Nature of Matter

Two identical photocathodes receive the light of frequencies f _ 1...

Two identical photocathodes receive the light of frequencies f1{f}_{1} and f2{f}_{2} respectively. If the velocities of the photo-electrons coming out are v1{v}_{1} and v2{v}_{2} respectively, then

Options

  1. Av12−v22=2hm[f1−f2]{v}_{1}^{2} - {v}_{2}^{2} = \dfrac{2h}{m}[ {f}_{1} - {f}_{2} ]
  2. Bv12+v22=2hm[f1+f2]{v}_{1}^{2} + {v}_{2}^{2} = \dfrac{2h}{m}\left[{f}_{1} + {f}_{2}\right]
  3. Cv1+v2={v}_{1} + {v}_{2} =\left[\dfrac{2h}{m}\left({f}_{1} + {f}_{2}\right)\right]\dfrac{1}{2}
  4. Dv1−v2=[2hm(f1−f2)]12{v}_{1} - {v}_{2} = [ \dfrac{2h}{m}\left({f}_{1} - {f}_{2}\right) ]\dfrac{1}{2}

Correct answer

A. v12−v22=2hm[f1−f2]{v}_{1}^{2} - {v}_{2}^{2} = \dfrac{2h}{m}[ {f}_{1} - {f}_{2} ]

Step-by-step solution

12mv12=hf1−ϕ\dfrac{1}{2}m{v}_{1}^{2} = h{f}_{1} - ϕ

12mv22=hf2−ϕ\dfrac{1}{2}m{v}_{2}^{2} = h{f}_{2} - ϕ

v12−v22=2hm(f1−f2){v}_{1}^{2} - {v}_{2}^{2} = \dfrac{2h}{m}\left({f}_{1} - {f}_{2}\right)

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