A small plate of photosensitive metal having work function $1.1eV$ is placed at a distance of $2m$from monochromatic light source of wavelength $496nm$ and power 1 watt. The light falls normally on the plate. Calculate the number of photons striking the metal plate per second per square meter. If a constant magnetic field of strength ${10}^{-4}T$ is applied parallel to the metal surface, calculate the radius of the largest circular path followed by the emitted photoelectron. (Given $hc=12400 \mathrm{eV}Å$)

 

Energy of each incident photon in $eV$ is 

$E=\frac{12400}{4960}eV=2.5eV$

$\Rightarrow E=2.5\times 1.6\times {10}^{-19}J=4\times {10}^{-19}J$

The rate of emission of photons i.e. the number of photons emitted per second by the source is

$n=\frac{P_{\text{source}}}{E}=\frac{1J{s}^{-1}}{4\times {10}^{-19}J}=2.5\times {10}^{18}\text{ photons }/s$

Hence, the number of photons striking the plate per second per square meter (also called as photon flux) is given by

${\varphi }_N=\frac{n}{A}=\frac{n}{4\pi r^2}=\frac{2.5\times {10}^{18}}{4(3.14)(2{)}^2}=5\times {10}^{16}{s}^{-1}{m}^{-2}$

The maximum kinetic energy of the photo-electrons emitted from the plate having work function ${\varphi }_0=1.1eV$ is given by

$K_{\max }=E-{\varphi }_0=2.5-1.1=1.4eV$

$\Rightarrow \frac{1}{2}m{v}_{\max }^2=1.4eV$

The maximum velocity of the ejected photoelectrons is

$v_{\max }=\sqrt{\frac{2\times 1.4\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}$

$\Rightarrow v_{\max }=7.0\times {10}^{5}m{s}^{-1}$

The maximum radius of the circle traversed by photoelectron in magnetic field $B={10}^{-4}T$ is given by

$r=\frac{mv}{qB}=\frac{\left(9.1\times {10}^{-31}\right)\left(7\times {10}^5\right)}{\left(1.6\times {10}^{-19}\right)\left({10}^{-4}\right)}=0.0398m$

$\Rightarrow r\cong 0.04\text{ metre }=4.0cm$