light of frequrency `7.5xx10^(14)` Hz is incident on a metal surface . A curve between `KE_(max) v//s` incident frequency is plotted as shown .Find
(a) Thee stopping potential
(b) Magnitude of the slope of curve given
(C ) Intercept on x-axis

Explanation

`KE_(max)=hv- hv _(0)" " ….(1)`
`hv_(0)=1.1eV` (from the given curve given )
Incident energy =hv
=`6.62xx10^(-34)xx7.5xx10^(-14)J` `rArr(6.62xx10^(-20)xx7.5)/(1.6xx10^(-19))eV` `rArr3.1eV`
`KE_(max)=3.1-1.1`=2eV
we know that
(a) stopping potential =`KE_(max)//e`
`=2V`
magnitude of slope of curve =h
`(6.62xx10^(-34)Js)/(1.6xx10^(-19))rArr4.14xx10^(-15)`
(c)Intercept on x-axis =Threshold frequency
`v_(0)=(1.1eV)/(heV-s)=(1.1)/(4.14xx10^(-15))=2.66xx10^(14)Hz`