A metal surface is illuminated by a radiation of wavelength 4500 A ∘ . The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90 ° with the magnetic field. If it starts

Question

A metal surface is illuminated by a radiation of wavelength 4500 A ∘ . The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90 ° with the magnetic field. If it starts revolving in a circular path of radius 2 mm , the work function of the metal is approximately

TestHub · Accepted Answer

From the Einstein's photoelectric equation $K = \frac{h c}{λ} - ϕ \Rightarrow ϕ = \frac{h c}{λ} - K$

Now $R = \frac{m v}{B q} \Rightarrow v = \frac{R B q}{m} \Rightarrow K = \frac{R^{2} B^{2} q^{2}}{2 m}$

Therefore, $ϕ = \frac{h c}{λ} - \frac{R^{2} B^{2} q^{2}}{2 m} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{4500 \times 10^{- 10}} - \frac{{(2 \times 10^{- 3} \times 2 \times 10^{- 3} \times 1.6 \times 10^{- 19})}^{2}}{2 \times 9.1 \times 10^{- 31}} \Rightarrow ϕ = (4.42 - 2.25) \times 10^{- 19} J = \frac{2.27}{1.6} eV = 1.36 eV$

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