A photon with energy of 4.9 eV ejects photoelectrons from tungsten. When the ejected electron enters a constant magnetic field of strength B = 2.5 mT at an angle of 60° with the field direction, the maximum pitch of the helix described by the electron is found to be 2.7 mm. Find the work-function of the metal in electron-volt. Given that, specific charge of electron is 1.76 $\times$ 10¹¹ C/ kg.

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answer is 4.5.

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Detailed Solution

Pitch of helical path, $p = (vcos θ) T = \frac{vT}{2} . (as θ = 60^{\circ})$

$T= \frac{2 πm}{Bq} = \frac{2 π}{Bα} (∵ α = \frac{q}{m})$

$\begin{array}{l} ∴ p = \frac{πv}{Bα} \\ or v = \frac{Bαp}{π} . . . . . . . . (i) \\ KE = \frac{1}{2} {mv}^{2} = E - W \\ ∴ W = E - \frac{1}{2} {mv}^{2} . . . . . . . . . (ii) \end{array}$

Substituting value of v from Eq. (i) in Eq. (ii),
we get

$\begin{array}{l} W = 4 .9 - \frac{1}{2} \\ 9 .1 \times 10^{- 31} \times {(2 .5 \times 10^{- 3})}^{2} {(1 .76 \times 10^{11})}^{2} \\ \times \frac{{(2 .7 \times 10^{- 3})}^{2}}{π^{2} \times 1 .6 \times 10^{- 19}} \\ = (4 .9 - 0 .4) eV = 4 .5 eV \end{array}$

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