Q. What is the strength of transverse magnetic field required to bend all the photoelectrons within a circle of a radius 50 cm. When light of wavelength 3800 $\mathring{\mathrm{A}}$ is incident on a barium emitter? (Given that work function of barium is $2.5\ eV;h = 6.63 \times 10^{- 34}$ J-s; $e = 1.6 \times 10^{- 19}$ C; $m = 9.1 \times 10^{- 31}$kg)
A) $6.32 \times 10^{- 4}$ T
B) $6.32 \times 10^{- 5}$ T
C) $6.32 \times 10^{- 6}$ T
D) $6.32 \times 10^{- 8}$ T
Explanation
(c) $\frac{1}{2}mv_{\max}^{2} = \left\lbrack \frac{hc}{\lambda} - \phi_{0} \right\rbrack$ or $v_{\max}^{2} = \frac{2}{m}\left\lbrack \frac{hc}{\lambda} - \phi_{0} \right\rbrack$
$= \frac{2}{9.1 \times 10^{- 31}} \times \left\lbrack \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{3.8 \times 10^{- 7}} - 2.5 \times 1.6 \times 10^{- 19} \right\rbrack$
$= 27.12 \times 10^{10}m^{2}s^{- 2}$
Now, $B = \frac{mv_{\max}}{eR_{\max}} = \frac{9.1 \times 10^{- 31} \times 5.21 \times 10^{5}}{1.6 \times 10^{- 19} \times 0.5}$
$= 6.32 \times 10^{- 6}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.