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Asked by bidyutpravarout79 | 26 Apr, 2024, 21:40: PM

Power of light source = 30 W = 30 J /s

Hence we get 30 J energy for every second .

If the radiation from light source is mainly 6600 A° , then energy E of each photon is

E = ( h c ) / λ , where h = 6.626 × 10-34 J s is planck's constant , c is speed of light and λ is wavelength of light .

E = ( 6.626 × 10-34 × 3 × 108) / ( 6600 ×10-10) J ≈ 3 × 10-19 J

Number of photons incident on cathode, n  = 30 / ( 3 × 10-19 ) = 1020

Let e = 1.602 × 10-19 C is charge on electron . Let k = 0.01 is efficiency of photo electric emission .

Hence current from photo cell is

I =k n e = 0.01 × 1020 × 1.602 × 10-19 = 0.16 A

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Question # 6

Saturation current is maximum current we get for the given intensity of radiation .

Maximum current is due to maximum number of photo electrons .

Maximum number of photo electrons emitted from cathode depends on intensity of radiation.

Intensity depends on distance from source to cathode .

Answered by Thiyagarajan K | 27 Apr, 2024, 10:02: AM
NEET neet - Physics
NEET neet - Physics
NEET neet - Physics
NEET neet - Physics
NEET neet - Physics