JEE Advanced 2022 Paper 2 Q16 Physics Modern Physics Photoelectric Effect Medium

JEE Advanced 2022 Paper 2 · Q16 · Photoelectric Effect

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0\,\text{V}$. This potential drops to $0.6\,\text{V}$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take $hc/e = 1.24\times 10^{-6}\,\text{J m C}^{-1}$.]

  1. A. $1.72\times 10^{-7}\,\text{m},\ 1.20\,\text{eV}$
  2. B. $1.72\times 10^{-7}\,\text{m},\ 5.60\,\text{eV}$
  3. C. $3.78\times 10^{-7}\,\text{m},\ 5.60\,\text{eV}$
  4. D. $3.78\times 10^{-7}\,\text{m},\ 1.20\,\text{eV}$
Reveal answer + step-by-step solution

Correct answer:A

Solution

Einstein's photoelectric equation: $eV_s = hc/\lambda - \phi$. For source 1 with wavelength $\lambda$: $6e = hc/\lambda - \phi$. For source 2 with $4\lambda$ (intensity doesn't change stopping potential): $0.6e = hc/(4\lambda) - \phi$. Subtract: $5.4 e = hc/\lambda - hc/(4\lambda) = (3hc)/(4\lambda)$, so $\lambda = 3hc/(4\cdot 5.4 e) = (3\cdot 1.24\times 10^{-6})/(21.6) = 3.72\times 10^{-6}/21.6 \approx 1.72\times 10^{-7}\,\text{m}$. Then $\phi = hc/\lambda - 6e = (1.24\times 10^{-6})/(1.72\times 10^{-7})\,\text{eV} - 6 = 7.21 - 6 \approx 1.20\,\text{eV}$. Answer: (A).

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