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The radiation emitted when an electron jumps from n=3 to n = 2 orbit in a hydrogen atom, falls on a metal to produce photoelectron. The electron with maximum kinetic energy are made to move perpendicular to a magnetic field of (1)/(320) T in a radius of 10^(m). The work function of the metal is : |
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Answer» 3.03 eV `(1)/(lambda)=R((1)/(2^(2)))-(1)/(3^(2))=(5R)/(36)` Given `r=10^(-3) m and B=(1)/(320)T` Now `r=(mv)/(EB)=(sqrt(2mk))/(eB)` Here K is K.E. of emitted electron `:.K=(r^(2)e^(2)B^(2))/(2m)` `=(10^(-6)xx(1.6xx10^(-19))^(2)XX1)/(2XX(9.1xx10^(-31))^(2)(320)^(2))=(10^(-17))/(8xx9.1)` `=(10^(-17))/(72.8xx1.6xx10^(-19))eV=0.86eV` Now `(HC)/(lambda)=K+w` `:.w=(hc)/(lambda)-K` `=(5Rhc)/(36)-K` `=(5xx13.6)/(36)eV=0.86eV` ` w=103.eV` is work FUNCTION of metal |
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