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ProblemsProblem-1Determine the work function of a metal in Joules if the maximum threshold wavelength is 1.10x10-7

m.Problem-2Determine the maximum speed of electrons emitted from a zinc surface if they are stopped by a 16 N/C uniform electric field over a distance of 3.0cm.Problem-3Problem # 6/p.396 (textbook).

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Problem-1

Solution:

Ø = hfc = hc/λc (c = λf)

Ø = (6.63x10-34)(3x108)/ (1.10x10-7)Ø =1.81x10-18 J.

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Problem -2

Solution:

• Calculate the voltage:

E = Vs /d Vs = E d = (16)(0.030)

Vs = 0.48 V

• Maximum kinetic energy of an emitted electron:

Ek max = q Vs

Ek max = (1.60× 10-19)(0.48) = 7.7×10-20 J

• Calculate the maximum speed of electrons:

Ek max = (½)mv2 v = sqrt(2 Ek max /m)

v = sqrt((2)(7.7×10-20)/ 9.11×10-31 = 4.11×105m/s.

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Problem-3 (Prob. 6/p.396)

Solution:a) Energy rate absorbed per electron:

Pe = P A = (5x10-4) (1.0x10-18) = 5x10-22 W (J/s)

Pe = Ø /Δt Δt = Ø /P = (3.0)(1.6x10-19)/5x10-22

Δt = 960 s = 16 min.

b) This implies that the electrons are NOT being emitted almost

instantaneously, which contradicts the photoelectric effect.Calculate the critical frequency of the metal:

Ø = hfc fc = Ø/h = (3.0)(1.6x10-19)/ 6.63x10-34

= 7.24x1014 Hz.

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Problem-3, cont’d

Calculate the frequency of the incident radiation:E = hf f = E/h = Pe Δt/h = (5x10-22 J/s)(1 s)/ 6.63x10-34

f = 7.54 x 1011 Hz < fc no emission will take place.

c) Energy of the emitted electrons depend only on light frequency. The photoelectric effect shows that the current flows almost immediately as light falls on the metal surface, independently of light intensity, as long as f > fc .