Light of wavelength 180 nm ejects photoelectron from a plate of a metal whose work function is 2 eV. If a uniform magnetic field of 5×10−5T is applied parallel to plate, what would be the radius (in metre) of the path followed by electrons ejected normally from the plate with maximum energy?

If vmax is the speed of the fastest electron emitted from the metal surface, thenhcλ=W0+12mvmax2⇒(6.63×10−34)×(3×108)(180×10−9)=2×(1.6×10−19)+12(9.1×10−31)vmax2⇒v=1.31×106m/s The radius of the electron is given by r=mvqB=(9.1×10−31)×(1.31×106)(1.6×10−19)×(5×10−5)=0.149m

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Light of wavelength 180 nm ejects photoelectron from a plate of a metal whose work function is 2 eV. If a uniform magnetic field of $5 \times 10^{- 5} T$ is applied parallel to plate, what would be the radius (in metre) of the path followed by electrons ejected normally from the plate with maximum energy?

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answer is 0.149.

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Detailed Solution

If $v_{m a x}$ is the speed of the fastest electron emitted from the metal surface, then

$\begin{array}{l} \frac{hc}{λ} = W_{0} + \frac{1}{2} {mv}_{\max}^{2} \\ \Rightarrow \frac{(6.63 \times 10^{- 34}) \times (3 \times 10^{8})}{(180 \times 10^{- 9})} = 2 \times (1.6 \times 10^{- 19}) + \frac{1}{2} (9.1 \times 10^{- 31}) v_{\max}^{2} \\ \Rightarrow v = 1.31 \times 10^{6} m / s \\ The radius of the electron is given by \\ r = \frac{mv}{qB} = \frac{(9.1 \times 10^{- 31}) \times (1.31 \times 10^{6})}{(1.6 \times 10^{- 19}) \times (5 \times 10^{- 5})} = 0.149 m \end{array}$