Question

has an intensity of 0.035 W/m2. The electrons are photoelectrically emitted with a maximum speed of 6.1x105 m/s. Assume the absorption of every photon ejects an electron. (a) Convert the work function 2.89 eV to Joules. b) Determine the kinetic energy of the electron with a speed of 6.1x105 m/s. The mass of an electron is 9.11x10-31 kg (c) Determine the total energy required to both eject the electron and give it a velocity of 6.1x105 m/s. (d) Determine the total energy deposited on a 1 cm2 area (1.0x10-4 m2) in 1.00 s with light that has an intensity of 0.035 W/m2.

Answer 1

part A :

Work Function W0 = 2.89 eV

1eV = 1.6 *10^-19 J

so

Wo = 2.89 *1.6*10^-19

Wo = 4.624*10^-19 Joules

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part B:

Kinetic energy K = 0.5 mv^2

K =0.5 * 9.11*10^-31* (6.1*10^5)^2

K = 1.695*10^-19 J

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part C:

Total Energy = Wo + K

E = (4.624 + 1.695) *10^-19

E = 6.319 *10^-19 J

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part D :

Intensity I= power/area

Power = I A

P = 0.035 * 1*10^-4

P = Energy/time = 3.5 *10^-6

Energy E = 3.5*10^-6 * 1

E = 3.5*10^-6 Joules