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Ultraviolet light of wavelength 300 nm and intensity 1.0 Wm-2 falls on the surface of a photosensitive material. If one per cent of the incident photons produce photo-electrons, then the number of photoelectrons emitted per second from an area of 1.0 cm2 of the surface is nearly
 
  • a)
    9.61 x 1014 s-1
  • b)
    4.12 x 1013 s-1
  • c)
    1.51 x l012 s-1
  • d)
    2.13 x 1011 s-1
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Ultraviolet light of wavelength 300 nm and intensity 1.0 Wm-2 falls on...

Calculation of Number of Photoelectrons Emitted per Second:

Given data:
- Wavelength of ultraviolet light (λ) = 300 nm = 300 x 10^-9 m
- Intensity of light (I) = 1.0 Wm^-2
- Percentage of incident photons producing photo-electrons = 1%
- Area of surface (A) = 1.0 cm^2 = 1.0 x 10^-4 m^2

Calculations:
1. Energy of a photon (E) can be calculated using the formula: E = hc/λ, where h = Planck's constant (6.626 x 10^-34 Js) and c = speed of light (3 x 10^8 ms^-1).
2. Number of photons falling on the surface per second (N) can be calculated using the formula: N = I/E, where I = intensity of light.
3. Number of photoelectrons emitted per second (n) can be calculated using the formula: n = N x (Percentage of incident photons producing photo-electrons).
4. Finally, the number of photoelectrons emitted per second from the given area can be calculated using the formula: Number of photoelectrons emitted per second = n/A.

Substitute the values:
- Calculate the energy of a photon (E).
- Calculate the number of photons falling on the surface per second (N).
- Calculate the number of photoelectrons emitted per second (n).
- Calculate the final result for the number of photoelectrons emitted per second.

Therefore, the number of photoelectrons emitted per second from an area of 1.0 cm^2 of the surface is approximately 1.51 x 10^12 s^-1.
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Community Answer
Ultraviolet light of wavelength 300 nm and intensity 1.0 Wm-2 falls on...
Use formula
N = IA/E
must change cm^2 to m^2.
you will get this answer.
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Ultraviolet light of wavelength 300 nm and intensity 1.0 Wm-2 falls on the surface of a photosensitive material. If one per cent of the incident photons produce photo-electrons, then the number of photoelectrons emitted per second from an area of 1.0 cm2 of the surface is nearlya)9.61 x 1014 s-1b)4.12 x 1013 s-1c)1.51 x l012 s-1d)2.13 x 1011 s-1Correct answer is option 'C'. Can you explain this answer?
Question Description
Ultraviolet light of wavelength 300 nm and intensity 1.0 Wm-2 falls on the surface of a photosensitive material. If one per cent of the incident photons produce photo-electrons, then the number of photoelectrons emitted per second from an area of 1.0 cm2 of the surface is nearlya)9.61 x 1014 s-1b)4.12 x 1013 s-1c)1.51 x l012 s-1d)2.13 x 1011 s-1Correct answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Ultraviolet light of wavelength 300 nm and intensity 1.0 Wm-2 falls on the surface of a photosensitive material. If one per cent of the incident photons produce photo-electrons, then the number of photoelectrons emitted per second from an area of 1.0 cm2 of the surface is nearlya)9.61 x 1014 s-1b)4.12 x 1013 s-1c)1.51 x l012 s-1d)2.13 x 1011 s-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ultraviolet light of wavelength 300 nm and intensity 1.0 Wm-2 falls on the surface of a photosensitive material. If one per cent of the incident photons produce photo-electrons, then the number of photoelectrons emitted per second from an area of 1.0 cm2 of the surface is nearlya)9.61 x 1014 s-1b)4.12 x 1013 s-1c)1.51 x l012 s-1d)2.13 x 1011 s-1Correct answer is option 'C'. Can you explain this answer?.
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