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 of barium in eV? Given `h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C, m_e=9.1xx10^(-31)kg.`

Answer» Here,
`lambda=4000xx10^(-10)m=4xx10^-7`,
`r=50cm =0.50m, B=5.26xx10^-6T`.
Force on electron in magnetic field is,
`F=Bev=(mv^2)/r or v=(Ber)/m`
Work function of barium is given by
`phi_0=(hc)/lambda-1/2mv^2=(hc)/lambda-1/2mxx((Ber)/(m))^2`
`=(hc)/lambda-1/2 (B^2e^2r^2)/m`
`=((6.6xx10^(-34))xx(3xx10^8))/((4xx10^-7)xx(1.6xx10^(-19))`
`-1/2 xx((5.26xx10^-6)^2xx(1.6xx10^(-19))^2xx(0.50)^2)/((9.1xx10^(-31))xx(1.6xx10^(-19)))`
`~~2.5 eV`


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