Verified Solution • 2025–26 Session moderate

QJEEM21Ph0612: JEE Main 2021 Physics - Class 12 JEEM21Ph0612 Solution: Two identical photocathodes receive the light of frequencies f…

Two identical photocathodes receive the light of frequencies f1 and f2 respectively. If the velocities of the photo-electrons coming out are v1 and v2 respectively, then

Multiple Choice Options

  • A v1−v2=[2hm(f1−f2)]12{v_1} - {v_2} = {\left[ {{{2h} \over m}({f_1} - {f_2})} \right]^{{1 \over 2}}}
  • B v12+v22=2hm[f1+f2]v_1^2 + v_2^2 = {{2h} \over m}[{f_1} + {f_2}]
  • C v1+v2=[2hm(f1+f2)]12{v_1} + {v_2} = {\left[ {{{2h} \over m}({f_1} + {f_2})} \right]^{{1 \over 2}}}
  • D v12−v22=2hm[f1−f2]v_1^2 - v_2^2 = {{2h} \over m}[{f_1} - {f_2}]

Step-by-Step Verified Solution

12mv12=hf1−ϕ{1 \over 2}mv_1^2 = h{f_1} - \phi ___________(1)

12mv22=hf2−ϕ{1 \over 2}mv_2^2 = h{f_2} - \phi ___________(2)

Subtracting equation (1) by equation (2)

12mv12−12mv22=hf1−hf2{1 \over 2}mv_1^2 - {1 \over 2}mv_2^2 = h{f_1} - h{f_2}

v12−v22=2hm(f1−f2)v_1^2 - v_2^2 = {{2h} \over m}({f_1} - {f_2})

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