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A 100 watt sodium lamp is radiating light of wavelength 5890°A uniformly in all directions . At what distance the average density is 1 photon/cm^3

A 100 watt sodium lamp is radiating light of wavelength 5890°A uniformly in all directions .         At what distance the average density is 1 photon/cm^3

Grade:12

2 Answers

Arun
25763 Points
2 years ago

Power of the sodium lamp, P = 100 W

Wavelength of the emitted sodium light, λ = 589 nm = 589 × 10−9 m

Planck’s constant, h = 6.626 × 10−34 Js

Speed of light, c = 3 × 108 m/s

(a) The energy per photon associated with the sodium light is given as: E = hc/λ = 6.626 x 10-34 x 3 x 108 / 589 x 10-9 = 3.37 x 10-19 J

= 3.37 x 10-19 / 1.6 x 10-19 = 2.11 eV

 

(b) Number of photons delivered to the sphere = n.

The equation for power can be written as: P = nE

∴ n = P/E = 100 / 3.37 x 10-19 = 2.96 x 1020 photons/s

Therefore, every second, 2.96 x 1020 photons are delivered to the sphere

Khimraj
3007 Points
2 years ago
The energy of photon is given by E=hυ=hcλ=1990×10−28JλE=hυ=hcλ=1990×10-28Jλ =1990×10−285890A=3376×10−22J=1990×10-285890A=3376×10-22J ..(i) Given that the lamp is emitting energy at the rate of 100Js−1100Js-1 (power=100 W). Hence, number of photons N emitted is given by N=1003376×10−22≅3×1020N=1003376×10-22≅3×1020 photons s−1s-1 ..(ii) b. We regard the lamp as a point source. Therefore, at a distance r from the lamp, the light energy is uniformly distributed over the surface of sphere of radius r. So, N photons are crossing area 4πr24πr2 of spherical surface per second. So, flux at a distance r is given by n=N4πr2n=N4πr2 or r=(N4π)‾‾‾‾‾‾√r=(N4π) or r=(3×10204×3.14)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√cm= 488860km

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