Dual Nature of Matter

JEE Main level, single correct, medium

173 of 200

Question 173 of 200

The electric field of light wave is given as \vec{E}={10}^{–3}\cos ⁡\left(\dfrac{2\pi x}{5\times {10}^{–7}}-2\pi \times 6\times {10}^{14}t\right)x^{^}\dfrac{N}{C} . This light falls on a metal plate of work function 2eV2eV . The stopping potential of the photo-electrons is:
Given, EE (in eVeV ) =12375λ(inA˚)=\dfrac{12375}{\lambda (inÅ)}

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Answer: (D) 0.48V0.48V

E⃗=10−3cos[2πx5×10−7−2π×6×1014t]x^N/c\vec{E}={10}^{-3}cos\left[\dfrac{2\pi x}{5\times {10}^{-7}}-2\pi \times 6\times {10}^{14}t\right]\hat{x} N/c
k=2π5×10−7k=\dfrac{2\pi }{5\times {10}^{-7}}
And ω=6×1014×2π\omega =6\times {10}^{14}\times 2\pi ⇒f=6×1014Hz\Rightarrow f=6\times {10}^{14} Hz
E=ϕ+KEmaxE=\phi +K{E}_{max}
KEmax=hf−ϕK{E}_{max}=hf-\phi
=6.6×10−34×6×10141.6×10−19−2=\dfrac{6.6\times {10}^{-34}\times 6\times {10}^{14}}{1.6\times {10}^{-19}}-2
=2.475−2=0.48eV=2.475-2=0.48eV
∴∴ Stopping potential is 0.48V0.48 V

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