In a photoelectric effect experiment, a metallic surface of work function 2.2 eV is illuminated with a light of wavelength 400 nm. Assume that an electron makes two collisions before being emitted and in each collision 10% additional energy is lost.
Q. Find the kinetic energy of this electron as it comes out of the metal.

Options

A. 0.46 eV
B. 0.31 eV
C. 0.23 eV
D. none of these

Answer

B

Explanation

Energy of photon `E=(hc)/(lamda)=(1.24xx10^(3))/(400)=3.1 eV`
Remaining energy `=3.1-0.31=2.79eV`
Energy lost is first collision is
`(3.1)xx((10)/(100))=0.31eV`
Remaining energy is
`3.1-0.31=2.79 eV`
Energy lost in second collision is
`(2.79)xx((10)/(100))=0.279 eV`
Total energy lost two collisions is
`(0.31)+(0.279)eV=0.589 eV` So, from conservation of energy, we have
`(hc)/(lamda)=phi+KE_(max)+` energy lost in two collision
`3.1=2.2+KE_(max)+0.589`
`KE_(max)=0.31eV`
Total energy after second collision is `(2.79-0.279)=2.511 eV`
Energy lost in third collision is
`2.511xx(10)/(100)=0.2511 eV`
Remaining energy `=(2.511-0.2155)=2.2500 eV`
Energy lost in fourth collision
`=(2.2599xx(10)/(100))=0.2259 eV`
Remaining energy `=(2.2599-0.2250)=2.034 eV`
After the fourth collision, the electron doen not have enough energy to overcome the work function, so it cannot come out.