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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
  • a)
    1 : 1
  • b)
    2 : 1
  • c)
    4 : 1
  • d)
    1 : 4
Correct answer is option 'B'. Can you explain this answer?
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The light of two different frequencies whose photons have energies 3.8...
Solution:

The maximum kinetic energy of an electron emitted from a metal surface is given by:

Kmax = hν - φ

where h is Planck's constant, ν is the frequency of the incident light, and φ is the work function of the metal.

For the first frequency, the maximum kinetic energy of the emitted electrons is:

Kmax1 = hν1 - φ = 3.8 eV - 0.6 eV = 3.2 eV

For the second frequency, the maximum kinetic energy of the emitted electrons is:

Kmax2 = hν2 - φ = 1.4 eV - 0.6 eV = 0.8 eV

The ratio of the maximum speeds of the emitted electrons for the two frequencies is given by:

vmax1 / vmax2 = √(Kmax1 / m) / √(Kmax2 / m) = √(Kmax1 / Kmax2)

where m is the mass of the electron.

Substituting the values of Kmax1 and Kmax2, we get:

vmax1 / vmax2 = √(3.2 eV / 0.8 eV) = √4 = 2

Therefore, the ratio of maximum speeds of emitted electrons for the two frequencies is 2 : 1.

Hence, the correct option is B.
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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 bea)1 : 1b)2 : 1c)4 : 1d)1 : 4Correct answer is option 'B'. Can you explain this answer?
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