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`        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.`
3 years ago

Jitender Pal
365 Points
```										Sol. Given
σ = 1 * 10^–9 cm^–2, W base 0 (C base s) = 1.9 eV, d = 20 cm = 0.20 m, λ = 400 nm
we know → Electric potential due to a charged plate = V = E * d
Where E → elelctric field due to the charged plate = σ/E base 0
d → Separation between the plates.
V = σ/E base 0 * d = 1 * 10^-9 * 20/8.85 * 10^-12 * 100 = 22.598 V = 22.6
V base 0e = hv – w base 0 = hc/λ – w base 0 = 4.14 * 10^-15 *3 *10^8/4 * 10^-7 – 1.9
= 3.105 – 1.9 =1.205 ev
or, V base 0 = 1.205 V
As V base 0 is much less than ‘V’
Hence the minimum energy required to reach the charged plate must be = 22.6 eV
For maximum KE, the V must be an accelerating one.
Hence max KE = V base 0 + V = 1.205 + 22.6 = 23.8005 ev

```
3 years ago
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