Question:
Radiation, with wavelength \(6561\,\mathrm{\AA}\) falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of \(3\times10^{-4}\) T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
Radiation, with wavelength \(6561\,\mathrm{\AA}\) falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of \(3\times10^{-4}\) T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
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The largest radius corresponds to the maximum kinetic energy: \(K_{max}=\frac{(eBr)^2}{2m}\); then \(\phi=\frac{hc}{\lambda}-K_{max}\).
Updated On: Nov 5, 2026
- 1.8 eV
- 0.8 eV
- 1.1 eV
- 1.6 eV
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The Correct Option is C
Solution and Explanation
For circular motion \(r=\frac{mv}{eB}\Rightarrow v_{max}=\frac{eBr}{m}\).
\(v_{max}=\frac{1.6\times10^{-19}\times3\times10^{-4}\times10^{-2}}{9.1\times10^{-31}}\approx5.27\times10^{5}\) m/s.
\(K_{max}=\frac{1}{2}mv_{max}^2=\frac{(4.8\times10^{-25})^2}{2\times9.1\times10^{-31}}\approx1.27\times10^{-19}\) J \(\approx0.79\) eV.
Photon energy \(=\frac{12375\ \mathrm{eV\,\AA}}{6561\ \mathrm{\AA}}\approx1.886\) eV.
\(\phi=1.886-0.79\approx1.1\) eV.
Answer: (3) 1.1 eV
\(v_{max}=\frac{1.6\times10^{-19}\times3\times10^{-4}\times10^{-2}}{9.1\times10^{-31}}\approx5.27\times10^{5}\) m/s.
\(K_{max}=\frac{1}{2}mv_{max}^2=\frac{(4.8\times10^{-25})^2}{2\times9.1\times10^{-31}}\approx1.27\times10^{-19}\) J \(\approx0.79\) eV.
Photon energy \(=\frac{12375\ \mathrm{eV\,\AA}}{6561\ \mathrm{\AA}}\approx1.886\) eV.
\(\phi=1.886-0.79\approx1.1\) eV.
Answer: (3) 1.1 eV
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