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NCERT Exemplar · Q19

Q.Consider a thin target (10−2 m10^{-2}\ \text{m} square, 10−3 m10^{-3}\ \text{m} thickness) of sodium, which produces a photocurrent of 100 μA100\ \mu\text{A} when a light of intensity 100 W/m2100\ \text{W/m}^2 (λ=660 nm\lambda = 660\ \text{nm}) falls on it. Find the probability that a photoelectron is produced when a photon strikes a sodium atom. [Take density of Na =0.97 kg/m3= 0.97\ \text{kg/m}^3.]

Punjab PsebLong· 3mImportance★★★★★
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The number of sodium atoms in the target (≈2.5×1018\approx 2.5\times10^{18}) far exceeds the number of photons arriving each second (≈3.3×1016\approx 3.3\times10^{16}), so effectively every incident photon strikes an atom. The probability of producing a photoelectron per photon-atom strike is the ratio of photoelectrons emitted per second to photons incident per second: P≈0.019P \approx 0.019 (about 2%2\%).

Solution

Photoelectrons emitted per second. The photocurrent is i=100 μA=10−4 Ai = 100\ \mu\text{A} = 10^{-4}\ \text{A}, so

ne=ie=10−41.6×10−19=6.25×1014 s−1.n_e = \frac{i}{e} = \frac{10^{-4}}{1.6\times10^{-19}} = 6.25\times10^{14}\ \text{s}^{-1}.

Photons incident per second. The power intercepted by the target of area A=(10−2)2=10−4 m2A = (10^{-2})^2 = 10^{-4}\ \text{m}^2 is

Pinc=(100 W/m2)(10−4 m2)=10−2 W.P_{\text{inc}} = (100\ \text{W/m}^2)(10^{-4}\ \text{m}^2) = 10^{-2}\ \text{W}.

Each photon carries

Eγ=hcλ=(6.62×10−34)(3×108)660×10−9=3.0×10−19 J,E_\gamma = \frac{hc}{\lambda} = \frac{(6.62\times10^{-34})(3\times10^8)}{660\times10^{-9}} = 3.0\times10^{-19}\ \text{J},

so

nγ=PincEγ=10−23.0×10−19=3.3×1016 s−1.n_\gamma = \frac{P_{\text{inc}}}{E_\gamma} = \frac{10^{-2}}{3.0\times10^{-19}} = 3.3\times10^{16}\ \text{s}^{-1}.

Number of sodium atoms in the target. The volume is V=A×t=10−4×10−3=10−7 m3V = A\times t = 10^{-4}\times10^{-3} = 10^{-7}\ \text{m}^3; with ρ=0.97 kg/m3\rho = 0.97\ \text{kg/m}^3 and molar mass M=23×10−3 kg/molM = 23\times10^{-3}\ \text{kg/mol}, …

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