Two identical, photocathodes receive light of frequencies f 1 and f 2 . If the velocities…

Physics · JEE Main · NTA ExamsAtoms 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
  1. \(v_{1}^{2}-v_{2}^{2}=\frac{2 h}{m}\left(f_{1}-f_{2}\right)\)
  2. \(\mathrm{v}_{1}+\mathrm{v}_{2}=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{f}_{1}+\mathrm{f}_{2}\right)\right]^{1 / 2}\)
  3. \(\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{f}_{1}+\mathrm{f}_{2}\right)\)
  4. \(\mathrm{v}_{1}-\mathrm{v}_{2}=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)\right]^{1 / 2}\)

Answer

(A) v_ 1 ^ 2 -v_ 2 ^ 2 = 2 h m (f_ 1 -f_ 2 )

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