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 $2.5\,T$ 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 $h = 6 . 6 \times 10^{- 34} J s$

Answer 1

Given that the electrons enter the magnetic field normally,
$\therefore \theta =90^\circ $
$\therefore $ Magnetic force acting on electrons will be,
$F=qvBsin\theta =qvB$
$\therefore v=\frac{F}{q B}$
$\text{ Force, }F=10^{- 8}\text{dyne }=10^{- 13}N$
$\therefore v=\frac{10^{- 13}}{1 . 6 \times 10^{- 19} \times 2 . 5}=2.5\times 10^{5}ms^{- 1}$
Also, by using Einstein's photoelectric equation, we know
$K.E_{max}=h\left(\nu - \nu_{0}\right)$
$\therefore \frac{1}{2}mv^{2}=h\left(\nu - \nu_{0}\right)$
$\therefore \frac{1}{2}\times 9.1\times \left(10\right)^{- 31}\times \left(2 . 5 \times \left(10\right)^{5}\right)^{2}=h\left(\nu - \nu_{0}\right)$
$\therefore 28.44\times \left(10\right)^{- 21}=h\left(\nu - \nu_{0}\right)$
$\therefore \left(\nu - \nu_{0}\right)=\frac{28 . 44 \times \left(10\right)^{- 21}}{6 . 6 \times \left(10\right)^{- 4}}$
$=4.3\times 10^{13}Hz$