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Photons of wavelength lambda are inciden...

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……………………

A

`(hc)/(lambda) - m_(e)+ (e^(2) B^(2)R^(2))/(2m_(e))`

B

`(hc)/(lambda) + 2m_(e) [(eBR)/(2m_(e))]^(2)`

C

`(hc)/(lambda) - m_(e) c^(2) - (e^(2)B^(2)R^(2))/(2m_(e))`

D

`(hc)/(lambda) - 2m_(e) [(eBR)/(2m_(e))]^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A

`(hc)/(lambda) - 2m_(e) [(eBR)/(2m_(e))]^(2)`
Magnetic lorertz force = C entripetal 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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