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Grade 11Modern Physics

A monochromatic point source radiating wavelength 6000 Å, with power 2 watt, an aperture A of diameter 0.1 m and a large screen SC are placed as shown in fig. A photo emissive detector D of surface area 0.5 cm2 is placed at the centre of the screen. The efficiency of the detector for the photoelectron generation per incident photon is 0.9.
  1. Calculate the photon flux at the centre of the screen and the photocurrent in the detector.
  2. If the concave lens L of focal length 0.6 m is inserted I the aperture as shown, find the new values of photon flux and photocurrent. Assume a uniform average transmission of 80% from the lens.
  3. If the work function of the photoemissive surface is 1 eV. Calculate the values of the stopping. Potential in the two cases (without and with the lens in the aperture).

Question image for A monochromatic point source radiating wavelength
Profile image of Radhika Batra
12 Years agoGrade 11
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1 Answer

Profile image of Kevin Nash
12 Years ago
Hello Student,
Please find the answer to your question
. Energy of one photon, E = hc / λ = (6.6 x 10-34) (3.0 x 108) / 6000 x 10-10
= 3.3 x 10-19 J
Power of the source is 2 W or 2 J/s. Therefore, number of photons emitting per second,
N1 = 2 / 3. 3 x 10-19 = 6.06 x 1018 / s
At distance 0.6 m, number of photons incident unit area per unit time:
n2 = n1 / 4π(0.6)2 = 1. 34 x 1018 / m2 / s
Area of aperture is,
S1 = π / 4 d2 (0.1)2 = 7.85 x 10-3 m2
∴ Total number of photons incident per unit time on the aperture,
N3 = n2 s1 = ( 1. 34 xc 1018) (7.85 x 10+-3) /s
= 1.052 x 1016 / s
The aperture will become new source of light.
Now these photons are further distributed in all directions. Hence, at the location of detector, photons incident per unit area unit time :
N4 = n3 / 4 π ( 6 – 0.6)2 = 1.052 x 1016 / 4 π (5.4)2
= 2. 87 x 1013 s-1 m-2
This is the photon flux at the centre of the screen. Area of detector is 0.5 cm2 or 0.5 x 10-4 m2 . Therefore, total number of photons incident on the detector per unit time:
n5 = (0.5 x 10-4) (2.87 x 1013 d) = 1.435 x 109 s-1
The efficiency of photoelectron generation is 0.9. Hence, total photoelectrons generated per unit time :
n6 = 0.9 n5 = 1.2915 x 109 s-1
or, photocurrent in the detector :
i = (e)n6 = (1.6 x 10-19)(1.2915 x 109) = 2.07 x 10-10 A
(b) Using the lens formula :
1/ v – 1/ -0.6 = 1/ - 0.6 or v = - 0.3 m
i.e, image of source (say S’, is formed at 0.3 m from the lens,)
Total number of photons incident per unit on the lens are still n3 or 1.052 x 1016s. 80% of it transmits to seconds medium Therefore, at a distance of 5.7 m from S’ number of photons incident per unit are per unit time will be :
N1 = (80 / 100) ( 1. 05 x 1016) / (4 π) (5.7)2
This is the photon flux at the detector.
New value of photocurrent is :
i = ( 2. 06 x 1013) (0.5 x 10-4) (0. 9) ( 1.6 x 10-19)
= 1.483 x 10-10 A
(c) For stopping potential
hc / λ = (EK)max + W = eV0 + W
∴ eV0 = hc / λ – W = 3.315 x 10-19 / 1.6 x 10-19 – 1 = 1.07eV
∴ V0 = 1.07 Volt
Note : The value of stopping potential is not affected by the presence of concave lens as it changes the intensity and not the frequency of photons. The stopping potential depends on the frequency of photons.
Thanks
Kevin Nash
askIITians Faculty

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