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IV. Numerical problems · Q8

Q.When a 6000 Å light falls on the cathode of a photo cell and produces photoemission, a stopping potential of 0.8 V is required to stop the emission of electrons. Determine

(i) the frequency of the light
(ii) the energy of the incident photon
(iii) the work function of the cathode material
(iv) the threshold frequency and
(v) the net energy of the electron after it leaves the surface.
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Step 1. (i) Frequency: ν=cλ=3×1086000×10−10=5×1014 Hz\nu=\dfrac{c}{\lambda}=\dfrac{3\times10^{8}}{6000\times10^{-10}}=5\times10^{14}\ \text{Hz}.

Step 2. (ii) Photon energy: E=hν=(6.626×10−34)(5×1014)≈3.313×10−19 J≈2.07 eVE=h\nu=(6.626\times10^{-34})(5\times10^{14})\approx3.313\times10^{-19}\ \text{J}\approx2.07\ \text{eV}.

Step 3. (iii) Work function: since Kmax=eV0=0.8K_{max}=eV_0=0.8 eV, ϕ0=E−Kmax=2.07−0.8≈1.27 eV\phi_0=E-K_{max}=2.07-0.8\approx1.27\ \text{eV}. …

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