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JEE Main 2026 (24 Jan, Shift 2) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2026 (24 Jan, Shift 2)

Q17JEE Main 2026 · 24 JAN · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is ______ m.

(h=6.63×10−34 J.s,e=1.6×10−19 C,c=3×108 m/s)\left(h = 6.63 \times 10^{-34}\text{ J.s}, e = 1.6 \times 10^{-19}\text{ C}, c = 3 \times 10^{8}\text{ m/s}\right)

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Answer: (b)

  1. hcλ−ϕ=3.2 eV\frac{hc}{\lambda} - \phi = 3.2\text{ eV} and hc2λ−ϕ=0.7 eV\frac{hc}{2\lambda} - \phi = 0.7\text{ eV}
  2. Subtract: hc2λ=2.5 eV⇒hcλ=5 eV\frac{hc}{2\lambda} = 2.5\text{ eV} \Rightarrow \frac{hc}{\lambda} = 5\text{ eV}
  3. λ=6.63×10−34×3×1085×1.6×10−19=2.49×10−7 m\lambda = \frac{6.63\times10^{-34}\times3\times10^{8}}{5\times1.6\times10^{-19}} = 2.49\times10^{-7}\text{ m}
  4. So the answer is (b) 2.5×10−72.5\times10^{-7} m.
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