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CBSE Class 12 Physics 2023 Outside Delhi Set 2 Paper

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Question : 7 of 13
Marks: +1, -0
A photon of wavelengths 663 nm663\,\text{nm} is incident on a metal surface. The work function of the metal is 1.50 eV1.50\,\text{eV}. The maximum kinetic energy of the emitted photo electrons is
Solution:     👈: Video Solution
hv=ϕ0+KEmax⁡h v = \phi_0 + K E_{\max}
or, hcλ=ϕ0+KEmax⁡\frac{h c}{\lambda} = \phi_0 + K E_{\max}
or, 6.62×10−34×3×108663×10−9\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{663 \times 10^{-9}} =1.5×1.6×10−19+KEmax⁡= 1.5 \times 1.6 \times 10^{-19} + KE_{\max}
∴KEmax⁡=0.6×10−19\therefore K E_{\max} = 0.6 \times 10^{-19}
=6×10−20 J= 6 \times 10^{-20} \,\text{J}
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