Light of wavelength `2000 Å` falls on an aluminium surface . In aluminium `4.2 e V` of energy is required to remove an electron from its surface. What is the kinetic energy , in electron volt of (a) the fastest and (b) the slowest emitted photo-electron . ( c) What is the stopping potential ? (d) What is the cut - off wavelength for aluminum? (Plank's constant `h = 6.6 xx 10^(-34) J-s` and speed of light `c = 3 xx 10^(8) m s^(-1).

Answer

(a) 2 eV (b) zero (c) 2V (d) 2946 Å

Explanation

Energy of the incident photon,
`E=(hc)/lambda=(6.6xx10^(-34)xx3xx10^(8))/(2000xx10^(-10))J=9.9xx10^(-19)J`
`=(9.9xx10^(-19))/(1.6xx10^(-19))eV~~6.2 eV`
(a) K.E. of the fastest electron,
`K=E-phi_(0)=6.2-4.2=2 eV`
(b) K.E. of the slowest electron=zero. As the emitted electrons have all possible energies from 0 to certain maximum value K
(c) If `V_(0)` is the stopping potential, then
`V_(0)=K//e=2 eV//e=2V`
(d) If `lambda_(0)` is the cut off wavelength for alluminium, then, `phi_(0)=(hc)/(lambda_(0))`
or `lambda_(0)=(hc)/(phi_(0))=(6.6xx10^(-34)xx3xx10^(8))/(4.2xx1.6xx10^(-19))`
`=2946.4xx10^(-10)m=2946.4 A`