Question

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 $3800 \, \mathring{A}$ 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. $4.88\times 10^{- 4}T$
• B. $32\times 10^{- 5}T$
• C. $6.32\times 10^{- 6}T$
• D. $7.6\times 10^{- 8}T$

Answer 1

Answer: (C)

$\frac{1}{2} m v_{\max }^{2}=\frac{h c}{\lambda}-\phi_{0}$ or $v_{\max }^{2}=\frac{2}{m} \frac{h c}{\lambda}-\Phi_{0}$
$=\frac{2}{9.1 \times 10^{-31}} \times \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3.8 \times 10^{-7}}-2.5 \times 1.6 \times 10^{-19}$
$=27.12 \times 10^{10} m ^{2} s ^{-2}$
NoW, $B=\frac{m v_{\max }}{e R_{\max }}=\frac{9.1 \times 10^{-31} \times 5.21 \times 10^{5}}{1.6 \times 10^{-19} \times 0.5}$
$=6.32 \times 10^{-6} T$