1.

Photons of wavelength lambda are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius R by a perpendicular magnetic field having magnitude B. The work function of the metal is……………………

Answer»

`(hc)/(lambda) - me_(e) + (e^(2) B^(2) R^(2))/(2m_(e))`
`(hc)/(lambda) + 2m_(e) [(eBR)/(2m_(e))]^(2)`
`(hc)/(lambda) - m_(e)c^(2) - (e^(2) B^(2) R^(2))/(2m_(e))`
`(hc)/(lambda) - 2m_(e) [(eBR)/(2m_(e))]^(2)`

Solution :Magnetic lorentz FORCE = Centripetal force
`BqV = (mv^(2))/(R)`
`therefore""V = (qBR)/(m) = (eBR)/(m)`
From Einstein.s photo electric equation
`KE_(max) = (hc)/(lambda) - phi`
`phi = (hc)/(lambda) - (1)/(2) m_(e) V^(2) = (hc)/(lambda) - (1)/(2) m_(e) ((eBR)/(m_(e)))^(2)`
`phi = (hc)/(lambda) - 2m_(e) ((eBR)/(2m_(e)))^(2)`


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