In a photoelectric effect experiment, the maximum kinetic energy of the emitted electrons is 1eV for incoming radiations of frequency `v_(0)` and 3eV for incoming radiation of frequency `3v_(0)//2`. What is the maximum kinetic energy of the electrons emitted for incomming radiation of frequency `9v_(0)//4`?

Options

A. `3 eV`
B. `4.5 eV`
C. `6 eV`
D. `9 eV`

Answer

c

Explanation

`(KE)_(max)=hv-phi_(0)`
So, `1 eV=hv_(0)-phi_(0)`......(i)
and `3eV=(h3v_(0))/2-phi_(0)`......(ii)
`3eV-1eV=hv_(0)//2` or `hv_(0)=4eV`
From (i), `phi_(0)=hv_(0)-1eV=4eV-1eV=3eV`
`:. (K.E.)_(max)=hxx(9v_(0))/4-3eV`
`=9/4(4eV)-3eV=6eV`