The normalized wave function for a hydrogen atom in the
1s state is given by
ψ(r) = where
α0 is the Bohr radius, which is equal to 5.29 × 10-11 m.
What is the probability of finding the electron at a distance
greater than 7.8 α0 from the proton?
Anwer is 2.3 × 10-5, but how can I get it?
probability distribution punction P = 4pi r^2* |ψ(r)|^2 = 4r^2/[a0^3] e^(-2r/a0)
probability = integral Pdr
= integral 4r^2/[a0^3] e^(-2r/a0) dr limit from 7.8a0 to infinity
= [a0^2+2a0x+2x^2]*-a0/[a0^3]*e^(-2r/a0) from r=7.8a0 to infinity
= [1+2*7.8 +2*7.8^2]* e^(-15.6)
= 2.3*10^-5 Answer
The normalized wave function for a hydrogen atom in the 1s state is given by ψ(r)...
The normalized wave function for a hydrogen atom in the 1s state is given by ψ(r) =( 1 /(\sqrt{\pi a_{0}}) )e^{-r/a_{0}} \) where α0 is the Bohr radius, which is equal to 5.29 × 10-11 m. What is the probability of finding the electron at a distance greater than 7.8 α0 from the proton?
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