Light of wavlength of `2000 A` falls on an aluminimum surface. In aluminium, `4.2eV` are 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-electrons. © What is the stopping potential? (Planck's constant `h=6.6xx10^(-34)J s`, and speed of light `c=3xx10^(8)ms^(-1))`

Explanation

Energy corresponding to inclined photon,
`n = 5` (forth excited state) to `n = 4` (third excited state) is
`h v = (h c)/lambda = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(2000 xx 10^(-10))`
`= 9.9 xx 10^(-19) J = (9.9 xx 10^(-19))/(1.6 xx 10^(-19)) e V = 6.2 e V`
a. The kinetic energy fastest electros,
`E_(k) = h v - W`
or `E_(k) = 6.2 e V - 4.2 e V = 2 e V`
b. The kinetic energy of slowest electron is zero, since the emitted electron have all posible energies from `0` to certain maximum value `E_(k)`
c. If `V_(s)` is the stopping potential, then `E_(k) = e V_(s)`
or `V_(s) = (E_(k))/(e) = (2 e V)/(e) = 2 V`
d. If `lambda_(0)` is the cut-off wavelength for alumnium, then `W = (h c// lambda_(0))`
or `lambda_(0) = (h c//W)`
`= (6.6 xx 10^(-34) xx 3 xx 10^(8))/(4.2 xx 1.6 xx 10^(-19)) = 3000 xx 10^(-10) m`
`= 3000Å`