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a) The energy of incident photon is given by h f = c / ${\lambda}$ = (6.626 × 10-34 × 3.0 × 108) / 1.50 × 10-9 = 1.325 × 10-18 J
K.Emax = h f – ${\phi}$ = (1.325 × 10-18) – (2 × 1.6 × 10-19) = 1.0 × 10-18 J (max. K.E of the emitted electrons)
But K.Emax = ½ mv2max.
Therefore;
1.0 × 10 -18 = ½ × 9.1 × 10-31 × V2max
V2max = (1.0 × 10-18 /9.1 × 10-31)1/2 = 1.5 × 106 m/s (max. velocity of emitted electrons).
b) ${\phi}$ = h fco and f o = ${\phi}$ / ${\lambda}$, ${\phi}$ = 2 × 1.6 × 10-19 fo = (2 × 1.6 × 10-19) / (6.626 × 10-34) = 4.8 × 1014 Hz (min. threshold frequency of the emitted electrons)
Mutiso answered the question on October 17, 2018 at 06:18
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