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Modern Physics — Hard JEE physics MCQ

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0 \mathrm{~V}$. This potential drops to $0.6 \mathrm{~V}$ 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 $\frac{h c}{e}=1.24 \times$ $10^{-6} \mathrm{~J} \mathrm{mC}^{-1}$.]
  1. A. $1.72 \times 10^{-7} \mathrm{~m}, 1.20 \mathrm{eV}$
  2. B. $1.72 \times 10^{-7} \mathrm{~m}, 5.60 \mathrm{eV}$
  3. C. $3.78 \times 10^{-7} \mathrm{~m}, 5.60 \mathrm{eV}$
  4. D. $3.78 \times 10^{-7} \mathrm{~m}, 1.20 \mathrm{eV}$

Solution

The correct option is **A**.

PHYSICS

hard • PYQ Reworded
Question
Read carefully, then pick the best option.
When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0 V6.0 \mathrm{~V}. This potential drops to 0.6 V0.6 \mathrm{~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 hce=1.24×\frac{h c}{e}=1.24 \times 10−6 JmC−110^{-6} \mathrm{~J} \mathrm{mC}^{-1}.]
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