A small piece of cesium metal `phi=1.9eV` is kept at a distance of 20 cm form a large metal plate having a charge density of `1.0xx 10^-9 C m^-2` on the surface facing the cesium piece. A monochromatic light of wavelength 400nm is incident on the cesium piece. Find the minimum and the maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.

Explanation

Given ` sigma= 1 xx 10^(-9) c / m^(2) ` ,
W_0 (Cs) = 1.9 eV `
` d = 20 cm = 0.20 m `,
` lambda = 400 nm`
We know , Electric potential due to a charged plate, `V = E xx d `
Where E elctric field due to the charged palt e` = sigma = E_0`
` `d = Separation between the plates.
` V = sigma / E_0 xx d `
` = 1 xx 10^(-9) xx 20 / 8.85 xx10^(-12)xx 100 `
`= 22.598 V = 22.6`
V_0 e = hv - W_0 = hc / lambda - W `
` = 4.14 xx 10^(-15) xx 3 xx 10^(8) / 4 xx 10^(7) - 1.9 `
` = 3.105 xx -1.9 = 1.205 eV`
or , `V_0 = 1.205 V `
As V_0 is much less than 'V' Hence the minimum energy required to reach the charged plate must be
` = 22.7 eV ` For maximum K.E the 'V' msut be an accelarating one. Hence maximum,
` K.E. = V_0 + V =1.205 + 22.6`
23.8005 e V `.