A photon with energy of 4.9 eV ejects photoelectrons from tungsten. When the ejcect phototelectrons from tungsten. When the ejected electron enters a cnstant magnetic field of strangth B=2.5 mT at an angle of `60^@` with the field direction, the maximum pitch fo 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 `1.76xx10^11 C/kg.

Answer

D

Explanation

Pitch of helical path, `P = (v cos theta) T = (vT)/(2)`
`(as theta = 60^@)`
`T = (2pim)/(Bq) = (2pi)/(Balpha) (alpha = (q)/(m))
`:. `P = (piv)/(Balpha)`
or `v = (Balphap)/(pi)….(i)`
`KE = (1)/(2)mv^2 = E -W`
`:. W = E -(1)/(2) mv^2 .....(ii)
`Substituting value of v from Eq. (i) in Eq. (ii)
we get
`W = 4.9 - (1)/(2)` `9.1xx10^(-31)xx(2.5xx10^(-3))^(2 )(1.76xx10^11)^2`
`xx(2.7xx10^(-3))^2/(pi^2xx1.6xx10^(-19)`
`=(4.9 - 0.4) eV = 4.5 eV`