The work function for caesium atom is `1.9 eV`. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength `500 nm`, calculate the kinetic energy and the velocity of the ejected photoelectron.

Explanation

It is given that the work function `(W_(0))` for caesium atom is 1.9 eV.
(a) From the `W_(0)=(hc)/(lambda_(0))` expression, we get:
`lambda_(0)=(hc)/(W_(0))`
Where, `lambda_(0)` = threshold wavelength h = Planck’s constant c = velocity of radiation Substituting the values in the given expression of `(lambda_(0))`:
`lambda_(0)=((6.623xx10Js)(3.0xx10^(8)ms^(-1)))/(1.9xx1.602xx10^(-19)J)`
`lambda_(0)=6.53xx10^(-7) m`
Hence, the threshold wavelength `lambda_(0)` is 653 nm.
(b) From the expression , `W_(0)=hv_(0)`, we get:
`v_(0)=(W_(0))/(h)`
Where, `ν_(0)` = threshold
frequency h = Planck’s
constant
Substituting the values in the given expression of `ν_(0)`:
`v_(0)=(1.9xx1.602xx10^(-19)J)/(6.26xx10^(-34)Js)`
`(1 eV=1.602xx10^(-19)J)v_(0)`
`=4.593xx10^(14)S^(-1)`
Hence, the threshold frequency of rediation `(v_(0))` is `4.593xx10^(14)s^(-1)`.
(c). According to the question`.:`
Wavelength used in irradiation `(lambda)` = 500 nm
Kinetic energy = h `(ν – ν_(0))`
`-hc((1)/(lambda)-(1)(lambda_(0)))`
`=(6.626xx10^(-34)Js)(3.0xx10^(8) ms^(-1))((lambda_(0)-lambda)/(lambda lambda_(0))`
`=(1.9878xx10^(-23)Jm)[((653-500)10^(-9)m)/((653)(500)10^(-18)m^(2))]`
`=((1.9878xx10^(-26))(153xx10^(9))/((653)(500))J`
`=9.3149xx10^(-20)J`
Kinetic energy of the ejected photoelectron `=9.3149xx10^(-20)J`
Since K.E. `=(1)/(2)mv^(2)=9.3149xx10^(-20)J`
`v=sqrt((2(9.3149xx10^(-20)J)/(9.10939xx10^(-31)))`
`=sqrt(2.0451xx10^(11m^(2)s^(-2))`
v=`4.52xx10^(5)ms^(-1)`
Hence, the velocity of the ejected photoelectron (v) is `4.52 xx 10^(5) ms^(–1)`.