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I Multiple Choice Questions · Q12

Q.A light of wavelength 500 nm is incident on a sensitive plate of photoelectric work function 1.235 eV. The kinetic energy of the photoelectrons emitted is (Take h=6.6×10−34h = 6.6 \times 10^{-34} Js):

(a) 0.58 eV
(b) 2.48 eV
(c) 1.24 eV
(d) 1.16 eV
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Step 1. The incident photon's energy is E=hcλE=\dfrac{hc}{\lambda}, numerically E=1240λ(nm) eV=1240500≈2.48 eVE=\dfrac{1240}{\lambda(\text{nm})}\ \text{eV}=\dfrac{1240}{500}\approx2.48\ \text{eV}.

Step 2. The photoelectrons' kinetic energy is Kmax=E−ϕ0=2.48−1.235≈1.24 eVK_{max}=E-\phi_0=2.48-1.235\approx1.24\ \text{eV}.

Step 3. This matches option (c). …

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