125k views
3 votes
Having integrated with respect to ϕ and θ, you now have the constant 4π in front of the integral and are left to deal with ∫[infinity]0A21(e−r/a)2r2dr=A21∫[infinity]0r2(e−r/a)2dr.

What is the value of A21∫[infinity]0r2(e−r/a)2dr?Express your answer in terms of A1 and a.
Find the unique positive value of A1.
Express your answer in terms of a and π.

User Aswathy S
by
4.3k points

1 Answer

1 vote

Answer:

Explanation:


\int\limits^(\infty)_0 {A^2_1} (e^(-r/a))r^2dr= {A^2_1}\int\limits^(\infty)_0r^2(e^(-r/a))^2\, dr)


=A_1^2\int\limits^(\infty)_0 r^2e^(-2r/a)\ dr


=A_1^2[(r^2e^(2r/a))/(-2/a) |_0^(\infty)-\int\limits^(\infty)_0 2r(e^(-2r/a))/(-2/a) \ dr]


=A^2_1[0+\int\limits^(\infty)_0 a\ r\ e^(-2r/a)\ dr]


=A^2_1[(a \ r \ e^(-2r/a))/(-2/a) |^(\infty)_0-\int\limits^(\infty)_0 (a \ e^(-2r/a))/(-2/a) \ dr]


=A_0^2[0-0+\int\limits^(\infty)_0 (a^2)/(2) e^(-2r/a)\ dr\\\\=A_1^2(a^2)/(2) \int\limits^(\infty)_0 e^(-2r/a)\ dr\\\\=A_1^2(a^2)/(2) [(e^(-2r/a))/(-2/a) ]^(\infty)_0


=(A_1^2a^2)/(2) -(a)/(2) [ \lim_(r \to \infty) [e^(-2r/a) -e^0]\\\\=(A_1^2a^2)/(2) -((a)/(2)) (0-1)


=(A_1^2a^3)/(4)


\therefore A_1^2\int\limits^(\infty)_0 r^2(e^(-r/a)) \ dr =(A_1^2a^3)/(4)

Find the unique positive value of A1


=4\pi ((A_1^2a^3)/(4) )\\\\=A_1^2a^3\pi\\\\A_1^2=(1)/(a^3\pi) \\\\A_1=\sqrt{(1)/(a^3\pi) }

User Abso
by
3.9k points