Verified Solution • 2025–26 Session moderate

QJEEM24Ph250e: JEE Main 2024 Physics - Class 12 JEEM24Ph250e Solution: When a metal surface is illuminated by light of wavelength ,…

When a metal surface is illuminated by light of wavelength λ\lambda , the stopping potential is 8 V8 \mathrm{~V} . When the same surface is illuminated by light of wavelength 3λ3 \lambda , stopping potential is 2 V2 \mathrm{~V} . The threshold wavelength for this surface is:

Multiple Choice Options

  • A 3 λ\lambda
  • B 9 λ\lambda
  • C 5 λ\lambda
  • D 4.5 λ\lambda

Step-by-Step Verified Solution

E=ϕ+Kmaxϕ=hcλ0 Kmax=eV08e=hcλ−hcλ0….. (i) 2e=hc3λ−hcλ0…... (ii)  on solving (i) & (ii) λ0=9λ

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