The work function for Cesium atom is 1.9 eV . Calculate the threshold frequency of the radiation . If the Cesium element is irradiated with a wavelength of 500 nm , calculate the kinetic energy of the ejected photoelectron .

Explanation

Case-1
Photo electric effect equation is
`hv=hv_(0)+1//2mv^(2)`
`w=hv_(0)`
`v_(0)=(w)/(h)`
`=(1.9xx1.602xx10^(-19))/(6.625xx10^(-34))`
`=(3.0438xx10^(-19))/(6.625xx10^(-34))=0.45944xx10^(15)`
`=4.5944xx10^(14)sec^(-1)`
`:.v_(0)=4.5944xx10^(14)sec^(-1)`
Case -II
Photo electric effect equation is
`E=(hc)/(lamda)`
`=(6.625xx10^(-34)xx3xx10^(8))/(5xx10^(-7))`
`=(19.878xx10^(-26))/(5xx10^(-7))`
`=3.9756xx10^(-19)J`
Kinetic Energy `(KE)=(1)/(2)mv^(2)`
From Photo electric effect
`=(1)/(2)mv^(2)=hv-hv_(0)`
K.E. `=3.9756xx10^(-19)-1.9xx1.602xx10^(-19)`
`=3.9756xx10^(-19)-3.0438xx10^(-19)=0.9318xx10^(-19)=9.318xx10^(-20)J`.
Given work function `hv_(0)=1.9ev`
`=1.9xx1.602xx10^(-19)J`.
`c=3xx10^(8)m//sec`
`lamda=500xx10^(-9)`
`=5xx10^(-7)m`.