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QJEEM22Ph3e33: JEE Main 2022 Physics - Class 12 JEEM22Ph3e33 Solution: The light of two different frequencies whose photons have…

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

Multiple Choice Options

  • A 1:01
  • B 2:01
  • C 4:01
  • D 1:04

Step-by-Step Verified Solution

3.8=0.6+12mv123.8 = 0.6 + {1 \over 2}mv_1^2

1.4=0.6+12mv221.4 = 0.6 + {1 \over 2}mv_2^2

⇒v12v22=3.20.8=41\Rightarrow {{v_1^2} \over {v_2^2}} = {{3.2} \over {0.8}} = {4 \over 1}

⇒v1v2=21\Rightarrow {{{v_1}} \over {{v_2}}} = {2 \over 1}

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