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MHT CET 2020 (19 Oct, Shift 2) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from MHT CET 2020 (19 Oct, Shift 2)

Q118MHT CET 2020 · 19 OCT · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation

If the maximum kinetic energy of emitted electrons in photoelectric effect is 3.2×10−19 J3.2 \times 10^{-19}\text{ J} and the work-function for metal is 6.63×10−19 J6.63 \times 10^{-19}\text{ J}, then stopping potential and threshold wavelength respectively are

[Planck's constant, h=6.63×10−34 Jh = 6.63 \times 10^{-34}\text{ J}-s]

[Velocity of light, c=3×108 msc = 3 \times 10^{8}\,\frac{\text{m}}{\text{s}} ]

[Charge on electron =1.6×10−19 C= 1.6 \times 10^{-19}\text{ C} ]

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Answer: (c)

  1. V0=Kmaxe=3.2×10−191.6×10−19=2 VV_0 = \frac{K_{max}}{e} = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2\text{ V}
  2. λ0=hcϕ=6.63×10−34×3×1086.63×10−19=3×10−7 m\lambda_0 = \frac{hc}{\phi} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6.63 \times 10^{-19}} = 3 \times 10^{-7}\text{ m}
  3. λ0=3000 A˚\lambda_0 = 3000\,\mathring{A}
  4. So the answer is (c) 2 V, 3000 A˚\mathring{A}.
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