Question

An electric light bulb is made of an evacuated, i.e. no air inside the bulb, glass bulb containing a long thin tungsten filament. The resistivity of tungsten is is 6.87x10 -7 Ohm meters at the temperature of the filament. The filament can be modeled as a 580 mm long thin cylinder with 0.046 mm diameter. The light bulb is plugged into a 120 V wall outlet. The power dissipated by the filament is radiated away as black body radiation with an emissivity of 0.205. Only 2% of the total light is radiated away in visible light. None of the light radiated comes in UV or higher frequencies on the electromagnetic spectrum. Assume that the light bulb radiates uniformly with equal intensities of light in all directions. A rough approximation is to assume that all of the visible light is emitted at the average wavelength of a visible photon, 560 nm. A solar panel is placed 3 meters away from the light bulb. This solar panel is 0.5 meters x 0.5 meters in area. The work function of the solar panel is 1.774 eV. The panel only absorbs 10% of the photons that strike the surface, the rest pass through or reflect away.

A. (2 pts.) What is the resistance of the filament?

B. (2 pts.) What is the power consumed by the light bulb?

C. (2 pts.) What is the temperature of the filament?

D. (2 pts.) What is the peak frequency of the light emitted by the filament? What portion of the electromagnetic spectrum is this?

E. (3 pts.) What is the intensity of the visible light received by the solar panel?

F. (1 pts.) What is the peak strength of the electric field received from visible light by the photoelectric panel?

G. (2 pts.) What is the acceleration of an electron in this field?

H. (3 pts.) How many photons eject electrons in the panel per second? Assume that all visible photons are 560 nm photons.

I. (2 pts.) What is the current produced in the panel?

J. (1 pts.) What is the average kinetic energy of the ejected electrons? Assume the average energy is the maximum energy of the ejected electron from the average photon

Answer 1

Note I am allowed to answer only 4 sub-parts

(a) , is the resistivity = 6.87 X 10^ (-7) Ohm mrs

L = length of filament = 580 mm = 580 X 10^(-3) mrs

A = area of cross section of filament = { 0.023 X 10^(-3) mrs}^2 X 3.14 = 1.66 X 10^(-9) m^2

R = 6.87 X 10^(-7) Ohm-m X 580 X 10^(-3) mrs / 1.66 X 10^(-9) m^2 = 239.8 = 240 Ohms

(b) Power consumed by bulb = 120^2 volts^2 / 240 = 60 Watts

(c) P = , where e = emissivity = 0.205 , A = area of surface of the filament = Pi X 0.046 X 10^(-3)X 580 X 10^(-3)

= stefan boltzman constant =5.67 X 10^(-8) Watts/m^2K^4

60 watts = 0.205 X pi X 0.046 X 10^(-3) X 580 X 10^(-3) X 5.67 X 10^(-8) T^4

T = [ 60 / (0.205 X pi X 0.046 X 10^(-3) X 580 X 10^(-3) X 5.67 X 10^(-8) ]^(1/4)

= 2801 deg K

(d) Peak freuency is given by

= 5.879 X 10^10 X 2801 deg K = 16517 X 10^10

= 1.66 X 10^14 Hz = 166 THz corresponding to Infra red Radiation