Exercise 1: The threshold of dark-adapted (scotopic) vision is 4.9 10 11 W/m 2 a
ID: 1495741 • Letter: E
Question
Exercise 1: The threshold of dark-adapted (scotopic) vision is 4.9 1011 W/m2 at a central wavelength of 500 nm. If light with this intensity and wavelength enters the eye when the pupil is open to its maximum diameter of 8.4 mm, how many photons per second enter the eye? ____________ photons/s
Exercise 2: Two light sources are used in a photoelectric experiment to determine the work function for a particular metal surface. When green light from a mercury lamp ( = 546.1 nm) is used, a stopping potential of 0.894 V reduces the photocurrent to zero.
(a) Based on this measurement, what is the work function for this metal? __________ eV (b) What stopping potential would be observed when using light from a red lamp ( = 642.0 nm)? _________ V
Explanation / Answer
1. Using E = hf
= h(c/lambda) = 6.63x10-34 3x108/500x10-9 = 3.98x10-19 J for the energy of each photon at this wavelength.
To turn the threshold which is a power/area into energy/sec
we have to multiply by the area of the pupil which is pi r2 = 3.14 (0.0084/2)2 = 5.538x10-5,
and then the energy/sec = 4.9x10-11*5.538x10-5 = 2.713x10-15 J/s.
This then is 2.713x10-15/3.98x10-19 = 6816.58 photons/sec............Ans.
2.
Einstein’s photoelectric effect equation is Kmax = hf , and the energy required to raise an electron through a 1 V potential is 1 eV, so that Kmax = eVs = 0.894 eV.
A photon from the mercury lamp has energy: hf = hc/
= 4.14 × 1015 eV s* 3.00 × 108 m/s / 546.1× 109 m
E = hf = 2.27 eV
hence by Eienstien's law E = Kmax + W , where W is the work funciton of the metal.
so that W = E - Kmax = 2.27 - 0.894 = 1.5366 ev = 1.376 eV...........Ans.
b) for the incident light E = 12400 / 6320 = 1.96 eV
hence, for the same metal , now Kmax = E - W = 1.96 - 0.894 = 1.066 eV ...........Ans.
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