Light of wavelength 5000 mathringA and intensity 39.8 W m -2 is incident on a metal surface. If only 1 % photons of incident light emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be nearly

Question

Light of wavelength 5000 mathringA and intensity 39.8 W m -2 is incident on a metal surface. If only 1 % photons of incident light emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be nearly (A) 1018 (B) 1020 (C) 1022 (D) 1024

Tardigrade Admin · Accepted Answer

Energy of photon E=(h c/λ)=(1242 eV - nm /500 nm )=2.484 eV =2.484 × 1.6 × 10-19 J =4 × 10-19 J text Intensity =39.8 ( W / m 2)=(39.8 J / s / m 2) Number of photons falling per second per unit area =(39.8/4 × 10-19( m 2 s ))(( J / J ))=1 × 1020 1 % of photon eject photoelectron =1 % × 1020=1018

Q. Light of wavelength $5000 \overset{˚}{A}$ and intensity $39.8 W m^{- 2}$ is incident on a metal surface. If only $1%$ photons of incident light emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be nearly

$1 0^{18}$

$1 0^{20}$

$1 0^{22}$

$1 0^{24}$

Q. Light of wavelength 5000A˚ and intensity 39.8Wm−2 is incident on a metal surface. If only 1% photons of incident light emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be nearly

1018

1020

1022

1024

Energy of photon E=λhc​=500nm1242eV−nm​=2.484eV =2.484×1.6×10−19J=4×10−19J Intensity =39.8m2W​=m239.8J/s​ Number of photons falling per second per unit area =4×10−19(m2s)39.8​(JJ​)=1×1020 1% of photon eject photoelectron =1%×1020=1018

Q. Light of wavelength $5000 \overset{˚}{A}$ and intensity $39.8 W m^{- 2}$ is incident on a metal surface. If only $1%$ photons of incident light emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be nearly

$1 0^{18}$

$1 0^{20}$

$1 0^{22}$

$1 0^{24}$