1.

The work function of a metal is `3.0eV`.It is illuminated by a light of wave length `3xx10^(7)m`.Calculate i) threshold frequency, ii)the maximum energy of photoelectrons, iii) the stopping potential. `(h=6.63xx10^(-34)Js` and `c=3xx10^(8)ms^(-1))`

Answer» i)`W=hnu_(0)=3.0eV=3xx1.6xx10^(-19)J`
Threshold frequency
`nu_(0)=W/h=(3xx1.6xx10^(-19))/(6.63xx10^(-34))=0.72xx10^(15)Hz`.
ii) Maximum kinetic energy `(k_(max))=h(nu-nu_(0))`
`lambda=3xx10^(-7)m,u=c/lambda=(3xx10^(8))/(3xx10^(-7))=1xx10^(15)Hz`
`K_(max)=h(nu-nu_(0))=6.63xx10^(-34)(1-0.72)xx10^(15)`
`J=1.8xx10^(-19)J`.
iii) `K_(max)=eV_(0)` where `V_(0)` is stopping potential in volt and `e` is the charge of electron
`V_(0)=K_(max)/r`. Here `K_(max)=1.86xx10^(-19)J` and
`e=1.6xx10^(-19)C, V_(0)=(1.86xx10^(-19)J)/(1.6xx10^(-19)C)=1.16V`


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