Light of wavelength 180 nm ejects photoelectrons from a plate of metal whose work function is 2 eV. If a uniform magnetic field of \( 5 \times 10^{-5}\,\mathrm{Wb/m^2} \) is applied parallel to the plate, the radius of the path followed by the electrons ejected normally from the plate with maximum energy is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.15 m
Einstein's photoelectric equation gives the maximum kinetic energy. An electron ejected normally moves perpendicular to the field (which is parallel to the plate), so it moves in a circle of radius \( r = mv/(eB) = \sqrt{2mK}/(eB) \).
\( \displaystyle E = \frac{hc}{\lambda} = \frac{1240\,\mathrm{eV\,nm}}{180\,\mathrm{nm}} = 6.89\,\mathrm{eV} \)
\( K_{\max} = 6.89 - 2 = 4.89\,\mathrm{eV} \)
\( K_{\max} = 4.89 \times 1.6\times10^{-19}\,\mathrm{J} = 7.82\times10^{-19}\,\mathrm{J} \)
\( \displaystyle p = \sqrt{2mK_{\max}} = \sqrt{2 \times 9.1\times10^{-31} \times 7.82\times10^{-19}} = 1.19\times10^{-24}\,\mathrm{kg\,m\,s^{-1}} \)
\( \displaystyle r = \frac{p}{eB} = \frac{1.19\times10^{-24}}{1.6\times10^{-19} \times 5\times10^{-5}} = \frac{1.19\times10^{-24}}{8\times10^{-24}} \approx 0.15\,\mathrm{m} \)