A photon with an energy of $4.9 e V$ ejects photoelectrons from tungsten. When the ejected electron enters a constant magnetic field of strength $B = 2.5 m T$ at an angle of $60 °$ with the field direction, the maximum pitch of the helix described by electron is found to be $2.7 m m$ . Find the work function of the metal in electron-volts. Given that specific charge of electron is $1.76 \times 10^{11} C k g^{- 1}$ .

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$4.5 e v$

$4 e v$

$6.5 e v$

$6 e v$

answer is B.

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Detailed Solution

Pitch of helical path is

$p = (v \cos θ) T = \frac{v T}{2}$

$\{∵ θ = 60 °\}$

where, $T = \frac{2 π m}{q B} = \frac{2 π}{B α}$ $\{∵ α = \frac{q}{m}\}$

$\Rightarrow p = \frac{π v}{B α}$

$\Rightarrow v = \frac{B α p}{π}$

$\Rightarrow v = \frac{(2.5 \times 10^{- 3}) (1.76 \times 10^{11}) (2.7 \times 10^{- 3})}{3.14}$

$\Rightarrow v = 3.78 \times 10^{5} {ms}^{- 1}$

Since, $K E = \frac{1}{2} m v^{2} = E - ϕ_{0}$

$\Rightarrow ϕ_{0} = E - \frac{1}{2} m v^{2}$

Substituting value of (v) from equation (1) in equation (2), we get

$ϕ_{0} = 4.9 - \frac{1}{2} \frac{(9.11 \times 10^{- 31}) {(3.78 \times 10^{5})}^{2}}{1.6 \times 10^{- 19}}$

$\Rightarrow ϕ_{0} = (4.9 - 0.4) e V$

$\Rightarrow ϕ_{0} = 4.5 e V$

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