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f(θ) = (-2m / (h/2π )2 K ) * integral 0 to ∞ (r sin Kr V(r) dr)

Substituting V (r) = -V0 exp(-r2/2a2)

We get f(θ) = (-mVoa3√π / 2(h/2π) ) * exp (-k2a2sin2θ/2 )

Differential crossection dσ / dΩ = / f(θ) /^2

= (-mVoa3√π / 2(h/2π) )2 * exp (-2k2a2sin2θ/2 )

can u plz give the intermediate calculations.