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The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons ? (b) stopping potential and (c) maximum speed of emitted electrons ? |
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Answer» Given (ϕ0 = 2.14 eV, v = 6 × 1014 Hz (a) Maximum kinetic energy of emitted electron Ek = hv – (ϕ0 = 6.63 × 10–34 × 6 × 1014 –2.14 × 1.6 × 10–19 = 0.54 × 10–19 J =(0.54 x 10-19/1.6 x 10-19 eV) = 0.34 eV (b) Stopping potential v0 is given by Ek = eV0 ⇒V0 = Ek/e = 0.34eV/e = 0.34V (c) Maximum speed (vmax ) of emitted electrons is given by 1/2 mv2max = Ek or vmax = √2Ek/m = √(2 x 0.54 x 10-19/9.1 x 10-31) = 3.44 x 105 m/s |
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