Intro Semiconductor Device Physics Question #2 (30 points): An engineer is desig

Question

Intro Semiconductor Device Physics

Question #2 (30 points): An engineer is designing a new sodium photocathode for a night vision scope. She uses the experiment shown below to measure the voltage and current from the metallic emitter as a function of both the power of the light and wavelength of the light that hits the emitter. She assumes that all the electrons emitted go to the collector. Fill out the tables below with the results the engineer expects to take to her manager to characterize the device her company is building. Metallic emitterLight Collector Sensitive ammeter R=1.00

Explanation / Answer

when power of light ,P = 0.1 mW, wavelength = 600 nm = lambda
then if n photons hit photocathode per second
n*hc/lambda = P
then 0.1*10^-3 = nhc/600*10^-9 [ h = 6.63*10^-34 (planks constant) , c is speed of light, 3*10^8 m/s]

putting in values, n = 3.0165*10^14 photons per second
expected current, i = n*e [ assuming every photon releases an electroin, e is charge on electron ]
i = n*1.6*10^-19 = 4.82*10^-5 A

Voltage, V = iR ( R = 1 ohm)
V = 4.82*10^-5 V

when power of light ,P = 1 mW, wavelength = 550 nm = lambda
then if n photons hit photocathode per second
n*hc/lambda = P
then 10^-3 = nhc/550*10^-9 [ h = 6.63*10^-34 (planks constant) , c is speed of light, 3*10^8 m/s]

putting in values, n = 27.65*10^14 photons per second
expected current, i = n*e [ assuming every photon releases an electroin, e is charge on electron ]
i = n*1.6*10^-19 = 4.42*10^-4 A

Voltage, V = iR ( R = 1 ohm)
V = 4.42*10^-4 V

when power of light ,P = 10 mW, wavelength = 450 nm = lambda
then if n photons hit photocathode per second
n*hc/lambda = P
then 10*10^-3 = nhc/450*10^-9 [ h = 6.63*10^-34 (planks constant) , c is speed of light, 3*10^8 m/s]

putting in values, n = 226.244*10^14 photons per second
expected current, i = n*e [ assuming every photon releases an electroin, e is charge on electron ]
i = n*1.6*10^-19 = 36.199*10^-4 A

Voltage, V = iR ( R = 1 ohm)
V = 36.199*10^-4 V

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