"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"

"${1 \\over 2}mv_1^2 = h{f_1} - \\phi$ ___________(1) ${1 \\over 2}mv_2^2 = h{f_2} - \\phi$ ___________(2) Subtracting equation (1) by equation (2) ${1 \\over 2}mv_1^2 - {1 \\over 2}mv_2^2 = h{f_1} - h{f_2}$ $v_1^2 - v_2^2 = {{2h} \\over m}({f_1} - {f_2})$"

Easy MCQ +4 / -1 PYQ · JEE Mains 2021

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

A ${v_1} - {v_2} = {\left[ {{{2h} \over m}({f_1} - {f_2})} \right]^{{1 \over 2}}}$
B $v_1^2 + v_2^2 = {{2h} \over m}[{f_1} + {f_2}]$
C ${v_1} + {v_2} = {\left[ {{{2h} \over m}({f_1} + {f_2})} \right]^{{1 \over 2}}}$
D $v_1^2 - v_2^2 = {{2h} \over m}[{f_1} - {f_2}]$ Correct answer

Solution

${1 \over 2}mv_1^2 = h{f_1} - \phi$ ___________(1) ${1 \over 2}mv_2^2 = h{f_2} - \phi$ ___________(2) Subtracting equation (1) by equation (2) ${1 \over 2}mv_1^2 - {1 \over 2}mv_2^2 = h{f_1} - h{f_2}$ $v_1^2 - v_2^2 = {{2h} \over m}({f_1} - {f_2})$

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Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect

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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

Solution

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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