Jitender Pal
Last Activity: 10 Years ago
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