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In a photoelectric experiment, a monochromatic light is incident on the emitter plate (E ), as shown in the figure. When switch (S₁ ) is closed and switch (S₂ ) is open, the photoelectrons strike the collector plate (C ) with a maximum kinetic energy of (1 eV ). If switch (S₁ ) is open and switch (S₂ ) is closed and the frequency of the incident light is doubled the photoelectrons strike the collector plate with a ma

Options

  1. A(5233.3 Å )
  2. B(4133.3 Å )
  3. C(4166.7 Å )
  4. D(5336.7 Å )

Correct answer

B. (4133.3 Å )

Step-by-step solution

Let threshold frequency of emitter plate (=v₀ ) Energy of photon in first case is ( E ). When switch (S₁ ) is closed and switch (S₂ ) is open, So, (E=h v₀+(5+1) eV ) ...(i) For second case, when switch (S₁ ) is open and switch (S₂ ) is closed and frequency of incident light is doubled. then, ( 2 E=h v₀+(20-5) eV ) ...(ii) From Eqs. (i) and (ii), we get ( aligned & 2 (h v₀+6 eV )=h v₀+15 eV & 2 h v₀+12 eV =h v₀+15 eV & h v₀=3 eV & v₀= 3 1.6 10⁻¹⁹ 6.62 10⁻³⁴ =7.25 10¹⁴ ~Hz & ₀= c v₀ = 3 10^8 7.25 10¹⁴ & ₀=41333 Å ali

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