1.

When barium is irradiated by a light of lamda=4000A all the photoelectrons emitted are bent in a circle of radius 50 cm by a magnetic field of flux density 5.26xx10^(-6) T acting perpendicular to plane of emission of photoelectron. Then,

Answer»

the kinetic energy of fastest photoelectrin is 0.6 eV
work FUNCTION of the metal is 2.5 eV
the maximum velocity of photoelectron is `0.46xx10^(6)ms^(-1)`
the stopping potential for photoelectric effect is 0.6 V

Solution :`(1)/(2)mv_(max)^(2)=v-W`
DUE to MAGNETIC field,
`(mv_(max)^(2))/(r)=BeV_(max)`
`impliesv_(max)=(BER)/(m)`
`=(5.26xx10^(-6)xx1.6xx10^(-19)xx0.5)/(9.1xx10^(-31))`
`=0.46xx10^(6)ms^(-1)`
`(KE)_(max)=(1)/(2)mv_(max)^(2)=(Bev_(max)r)/(2)`
`=(5.26xx10^(-6)xx1.6xx10^(-19)xx0.46xx10^(6)xx0.5)/(2)`
`=0.973xx10^(-19)J=0.6eV`
Energy of PROTON,
`E=(hc)/(lamda)=(1240eVnm)/(400nm)=3.1eV`
Work function, `W=3.1 eV-0.6eV=2.5 eV`
`(KE)_(max)=eV_0impliesV_0=0.6V`


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