Question

A metal surface of threshold frequency $10^{15}Hz$ is illuminated by a light of frequency $\nu \left(\nu >{\nu }_0\right)$ . If the magnetic force by a $2.5T$ magnetic field acting on most energetic electrons that normally entered the magnetic is $10^{-8}dyne$ , then the difference between the frequency of incident radiation and the threshold frequency is $4.3\times 10^{x}Hz$ . Find the value of $x$ . (Take $h=6.6\times 10^{-34}Js$

Answer 1

Answer: 13

Given that the electrons enter the magnetic field normally,
$\therefore \theta =90^{\circ }$
$\therefore$ Magnetic force acting on electrons will be,
$F=qvBsin\theta =qvB$
$\therefore v=\frac{F}{qB}$
$\text{ Force, }F=10^{-8}\text{dyne }=10^{-13}N$
$\therefore v=\frac{10^{-13}}{1.6\times 10^{-19}\times 2.5}=2.5\times 10^{5}m{s}^{-1}$
Also, by using Einstein's photoelectric equation, we know
$K.E_{max}=h\left(\nu -{\nu }_0\right)$
$\therefore \frac{1}{2}m{v}^2=h\left(\nu -{\nu }_0\right)$
$\therefore \frac{1}{2}\times 9.1\times {\left(10\right)}^{-31}\times {\left(2.5\times {\left(10\right)}^5\right)}^2=h\left(\nu -{\nu }_0\right)$
$\therefore 28.44\times {\left(10\right)}^{-21}=h\left(\nu -{\nu }_0\right)$
$\therefore \left(\nu -{\nu }_0\right)=\frac{28.44\times {\left(10\right)}^{-21}}{6.6\times {\left(10\right)}^{-4}}$
$=4.3\times 10^{13}Hz$