Question

The work fuction of caseium is 1.8 eV. Light of 4500Å is incident on it. Calculate (i) the maximum kinetic energy of the emitted photoelectrons (ii) maximum velocity of the emitted photoelectrons (iii) if the intensity of the incident light is doubled, then find the maximum kinetic energy of the emitted photoelectrons. Given, `h=6.63xx10^(-34)Js.m_(e)=9.1xx10^(-31)kg, c=3xx10^(8)ms^(-1)`

Answer 1

Correct Answer - `(i) 1.52xx10^(-19)J (ii) 5.78xx10^(5) ms^(-1) (iii) 1.52xx10^(-19)J`
Here, `phi_(0)=1.8eV, lambda=4500 A =4.5xx10^(-7)m`
(i) Max K.E. of emitted photoelectrons is
`K_(max)=(hc)/lambda-phi_(0)`
`=(6.63xx10^(-34)xx3xx10^(8))/(4.5xx10^(-7))-1.8xx1.6xx10^(-19)`
`=4.4xx10^(-19)-2.88xx10^(-19)=1.52xx10^(-19)J`
(ii) Max. velocity of emitted photoelectron
`v_(max)=sqrt((2K_(max))/m)=sqrt((2xx1.52xx10^(-19))/(9.1xx10^(-31)))`
`=5.78xx10^(5)ms^(-1)`
(iii) Kinetic energy of the emitted photoelectron is independent of the intensity of the incident light. Hence, if intensity of incident light is double the max K.E. of the emitted photoelectron remains unchanged `(=1.52xx10^(-19)J)`