Answer 8_4_4
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Gold has a work function of 4.9 eV. (a) What is this in joules? (b) What is the maximum kinetic energy that the photoelectrons have if the gold is illuminated by UV light of frequency 1.7 ´ 1015 Hz (c) What is the stopping voltage of these electrons? (d) How fast do the electrons travel as they leave the surface? (10) Data: h = 6.63 × 10-34 Js e = 1.6 ´ 10-19 C mass of electron = 9.11 ´ 10-31 kg |
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(a)
E = 4.9 eV ´
1.6 ´
10-19 J/eV = 7.84 ´
10-19 J (P) |
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(b)
Work out photon energy, E = hf = 6.63 × 10-34 Js ´
1.7 ´
1015 Hz (P) |
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=
11.2 ´
10-19 J (P) |
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Ek
= hf - F
= 11.2 ´
10-19 J - 7.84 ´ 10-19 J = 3.36 ´
10-19 J (P) |
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(c)
Ek = eVs Þ
Vs = Ek/e |
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Vs
= 3.36 ´
10-19 J ¸ 1.6 ´ 10-19 C (P) |
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=
2.1 V (P) |
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(d)
v2 = 2Ek/m = 2 ´ 3.36 ´ 10-19 J ¸
9.11 ´
10-31 kg = 7.38 ´
1011 m2 s-2(P) |
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v
= 8.6 ´
105 m/s (P) |