Verified Solution • 2025–26 Session moderate

QJEEAdv22Ph5968: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph5968 Solution: When light of a given wavelength is incident on a metallic…

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 10−6JmC−110^{-6} \mathrm{JmC}^{-1} .]

Multiple Choice Options

  • A 1.72×10−7 m,1.20eV1.72 \times 10^{-7} \mathrm{~m}, 1.20 \mathrm{eV}
  • B 1.72×10−7 m,5.60eV1.72 \times 10^{-7} \mathrm{~m}, 5.60 \mathrm{eV}
  • C 3.78×10−7 m,5.60eV3.78 \times 10^{-7} \mathrm{~m}, 5.60 \mathrm{eV}
  • D 3.78×10−7 m,1.20eV3.78 \times 10^{-7} \mathrm{~m}, 1.20 \mathrm{eV}

Step-by-Step Verified Solution

By Einstein's photoelectric equation, we have:

KE=hv−ϕ\mathrm{KE} =h v-\phi

hv−ϕ=6eVh v-\phi =6 \mathrm{eV}

hcλ−ϕ=6eV\frac{h c}{\lambda}-\phi =6 \mathrm{eV} .........(i)

hc4λ−ϕ=0.6eV\frac{h c}{4 \lambda}-\phi =0.6 \mathrm{eV}

3hc4λ=5.4eV\frac{3 h c}{4 \lambda} =5.4 \mathrm{eV}

⇒\Rightarrow λ=3hc5.4eV×4\lambda=\frac{3 h c}{5.4 \mathrm{eV} \times 4}

⇒\Rightarrow λ=3×1.24×10−64×5.4=1.72××−7 m\lambda =\frac{3 \times 1.24 \times 10^{-6}}{4 \times 5.4}=1.72 \times \times^{-7} \mathrm{~m} ........(ii)

From equation (i) and (ii), we conclude

hc1.72×10−7×11.6×10−19−ϕ=6eV\frac{h c}{1.72 \times 10^{-7}} \times \frac{1}{1.6 \times 10^{-19}}-\phi=6 \mathrm{eV}

on subsituting all the values we get

amp;ϕ=(7.27−6)amp;⇒ϕ=1.27eVamp;⇒ϕ=1.20eV\begin{aligned}& \phi=(7.27-6) \\\\&\Rightarrow \phi=1.27 \mathrm{eV} \\\\&\Rightarrow \phi=1.20 \mathrm{eV}\end{aligned}

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