1.

Light of wavelength 4000Å is incident on barium. Photoelectrons emitted describe a circle of radius 50 cm by a magnetic field of flux density `5.26xx10^-6 tesla`. What is the work function of barium in eV? Given `h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C, m_e=9.1xx10^(-31)kg.`

Answer» `v_"max"^2=2/m["hc"/lambda-phi_0]`
`=2/(9.1xx10^(-31))[((6.6xx10^(-34))xx(3xx10^8))/(4xx10^(-7))-2.5xx1.6xx10^(-19)]`
`=20.88xx10^10`
`therefore v_"max"=4.57xx10^5` m/s
`Bev_"max"=(mv_"max"^2)/r`
`rArr r=(mv_"max")/"Be"`
`=((9.1xx10^(-31))xx(4.57xx10^5))/(5.2xx10^(-6)xx1.6xx10^(-19))=1/2 m`


Discussion

No Comment Found

Related InterviewSolutions