Two identical, photocathodes receive light of frequencies f 1 and f 2 . If the velocities…
Physics · JEE Main · NTA Exams — Atoms and Nuclei
Two identical, photocathodes receive light of frequencies f1 and f2. If the velocities of the photoelectrons (of mass m) coming out are respectively v1 and v2, then
- \(v_{1}^{2}-v_{2}^{2}=\frac{2 h}{m}\left(f_{1}-f_{2}\right)\)
- \(\mathrm{v}_{1}+\mathrm{v}_{2}=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{f}_{1}+\mathrm{f}_{2}\right)\right]^{1 / 2}\)
- \(\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{f}_{1}+\mathrm{f}_{2}\right)\)
- \(\mathrm{v}_{1}-\mathrm{v}_{2}=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)\right]^{1 / 2}\)
Answer
(A) v_ 1 ^ 2 -v_ 2 ^ 2 = 2 h m (f_ 1 -f_ 2 )
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