d increase the frequency will increase the voltage of a phot

d) increase the frequency will increase the voltage of a phototube e) Increase the frequency will increase the work 3) A copper cathode phototube is exposed to light with a wavelength of 200 nm. What is the potential difference between anode and cathode? The work function for copper Wi 4.7 eV 4) If the distance between the cathode and anode plates is 0.02 m in problem 3), a) What is the electric field between the plates? b) How much work needs to be done to move an electron from the cathode to anode? 5) A copper cathode phototube is exposed to light. The voltage of the phototube battery is VoA 7.72 V. The work function for copper is W 4.7 eV a) What is the photon energy of the incoming light? b) What is the wavelength of the incoming light? c) What is the kinetic energy of an ejected electron when it arrives at the 113 of the distance between the cathode and anode at point P shown in d) the diagram? P? What is the velocity of the electron at point Anode Cathode c

Solution

Solution:

4 (a)

Potential difference = -1.51 V

Then the field between cathode and anode E = - dV/dr = 1.51/0.02 = 75.5 N/C

4 (b)

Work done to move an electron = eE.dr = 1.6 x 10^-19 x 75.5 x 2 x 10^-2 = 2.41 x 10^-19 Joule

5 (b)

Wavelength = hc/eV = 6.63 x 10^-34 x 3x 10⁸ / (1.6 x 10^-19 x 7.72) = 1.61 x 10^-7 meter = 161 nm

5 (a) Energy = 1242.3/ wavelength (nm) = 1242.3/161 = 7.7 eV

5 (c) Kinetic energy at P = work done by field to accelerate it to point P ( x =1/3*r) from cathode (x=0)

= qEx(x=1/3*r) - qEx0

= 8.0 x 10^-20 Joule

5 (d) 1/2* m v² = 8.0 x 10^-20

V = (16 x 10^-20/ 9.1 x 10^-31)^1/2 = 4.19 x 10⁵ m/s