Physics Easy 145× since 2002 Q6638 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 = [ 2 h m ( f 1 − f 2 ) ] 1 2 {v_1} - {v_2} = {\left[ {{{2h} \over m}({f_1} - {f_2})} \right]^{{1 \over 2}}} v 1 − v 2 = [ m 2 h ( f 1 − f 2 ) ] 2 1 B v 1 2 + v 2 2 = 2 h m [ f 1 + f 2 ] v_1^2 + v_2^2 = {{2h} \over m}[{f_1} + {f_2}] v 1 2 + v 2 2 = m 2 h [ f 1 + f 2 ] C v 1 + v 2 = [ 2 h m ( f 1 + f 2 ) ] 1 2 {v_1} + {v_2} = {\left[ {{{2h} \over m}({f_1} + {f_2})} \right]^{{1 \over 2}}} v 1 + v 2 = [ m 2 h ( f 1 + f 2 ) ] 2 1 D v 1 2 − v 2 2 = 2 h m [ f 1 − f 2 ] v_1^2 - v_2^2 = {{2h} \over m}[{f_1} - {f_2}] v 1 2 − v 2 2 = m 2 h [ f 1 − f 2 ] Check answer Skip
JEE Main 2021 · 17 Mar, Evening shift · Q6638· Dual Nature of Radiation