The photosensitive surface is receiving the light of wavelength 5000 overset° A at the rate of 10- 8 J s- 1. The number of photons received per second is ( h=6.62× 10- 34Js,c=3× 108ms- 1 )

Question

The photosensitive surface is receiving the light of wavelength 5000 overset° A at the rate of 10- 8 J s- 1. The number of photons received per second is ( h=6.62× 10- 34Js,c=3× 108ms- 1 ) (A) 2.5× 1010 (B) 2.5× 1011 (C) 2.5× 1012 (D) 2.5× 109

Tardigrade Admin · Accepted Answer

Energy of photon E=(h c/λ ) Given, λ =5000 textâ« = 5× 10- 7 textm ∴ E=(6.6 × 10- 34 × 3 × 108/5 × 10- 7) = 3.96× 10- 19 textJ Energy received per second = 10- 8 J s- 1 ∴ Number of photon's received per second = (E n e r g y r e c e i v e d p e r s e c o n d /E n e r g y o f o n e p h o t o n ) = (10- 8/3.96 × 10- 19)=2.5× 1010

Q. The photosensitive surface is receiving the light of wavelength $5000 A^{\circ}$ at the rate of $1 0^{- 8} J s^{- 1} .$ The number of photons received per second is ( $h = 6.62 \times 1 0^{- 34} J s, c = 3 \times 1 0^{8} m s^{- 1}$ )

$2.5 \times 1 0^{10}$

$2.5 \times 1 0^{11}$

$2.5 \times 1 0^{12}$

$2.5 \times 1 0^{9}$

Q. The photosensitive surface is receiving the light of wavelength 5000A∘ at the rate of 10−8Js−1. The number of photons received per second is ( h=6.62×10−34Js,c=3×108ms−1 )

2.5×1010

2.5×1011

2.5×1012

2.5×109

Energy of photon E=λhc​ Given, λ=5000Å=5×10−7m ∴E=5×10−76.6×10−34×3×108​ =3.96×10−19J Energy received per second = 10−8Js−1 ∴ Number of photon's received per second = EnergyofonephotonEnergyreceivedpersecond​ = 3.96×10−1910−8​=2.5×1010

Q. The photosensitive surface is receiving the light of wavelength $5000 A^{\circ}$ at the rate of $1 0^{- 8} J s^{- 1} .$ The number of photons received per second is ( $h = 6.62 \times 1 0^{- 34} J s, c = 3 \times 1 0^{8} m s^{- 1}$ )

$2.5 \times 1 0^{10}$

$2.5 \times 1 0^{11}$

$2.5 \times 1 0^{12}$

$2.5 \times 1 0^{9}$