Q. 314.0( 1 Vote )

# A small piece of cesium metal (ϕ = 1.9 eV) is kept at a distance of 20 cm from a large metal plate having a charge density of 1.0 × 10^{–9} 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 photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.

Answer :

Given, work function

Charge density

Distance d=20cm

Wavelength λ=400nm

We know that electric potential due to charged plate

is given by ,

where E is the electric field due to charged plate

and d is the distance between two plates

=

= 59V

Now, from Einstein’s photoelectric equation

⇒

Now putting the values of h, c, λ, Ф we get

⇒

⇒

⇒

∴

∵ ≪≪V

∴ Minimum kinetic energy required to reach the large

Metal =22.6eV

Next, maximum kinetic energy

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