Question

A small piece of cesium metal ($\varphi =1.9eV$) is kept at a distance of 20 cm from a large metal plate having a charge density of 1.0 $\times {10}^{-2}C{m}^{-2}$ on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and the maximum kinetic energy of the photo electrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.

Answer 1

Given $\sigma =1\times {10}^{-9}c/m^2$

$W_0\left(Cs\right)=1.9eV$

$d=20cm=0.20m,$

$\lambda =400nm$

We know, Electric potential due to a charged plate,

V = E $\times$ d

Where E electric field due to the charged plate

= $\frac{\sigma }{E_0}$

d = Separation between the plates.

$V=\frac{\sigma }{E_0}\times d$

$=\frac{1\times {10}^{-9}\times 20}{8.85\times {10}^{-12}\times 100}$

$=22.598V=22.6$

$V_{0}e=hv-W_0=\frac{hc}{\lambda }-W$

$=\frac{4.14\times {10}^{-15}\times 3\times {10}^8}{4\times {10}^7}-1.9$

= 3.105 - 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' must be an accelerating one.

Hence maximum, K.E. = $V_0+V=1.205+22.6$

= 23.8005 eV