Mathematics, 11.06.2020 17:57 claydale1659
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∫[in finity]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.
Answers: 3
![=A_1^2[\frac{r^2e^{2r/a}}{-2/a} |_0^{\infty}-\int\limits^{\infty}_0 2r\frac{e^{-2r/a}}{-2/a} \ dr]](/tpl/images/0683/0160/ee0ef.png)
![=A^2_1[0+\int\limits^{\infty}_0 a\ r\ e^{-2r/a}\ dr]](/tpl/images/0683/0160/d10f7.png)
![=A^2_1[\frac{a \ r \ e^{-2r/a}}{-2/a} |^{\infty}_0-\int\limits^{\infty}_0 \frac{a \ e^{-2r/a}}{-2/a} \ dr]](/tpl/images/0683/0160/aea11.png)
![=A_0^2[0-0+\int\limits^{\infty}_0 \frac{a^2}{2} e^{-2r/a}\ dr\\\\=A_1^2\frac{a^2}{2} \int\limits^{\infty}_0 e^{-2r/a}\ dr\\\\=A_1^2\frac{a^2}{2} [\frac{e^{-2r/a}}{-2/a} ]^{\infty}_0](/tpl/images/0683/0160/35fbb.png)
![=\frac{A_1^2a^2}{2} -\frac{a}{2} [ \lim_{r \to \infty} [e^{-2r/a} -e^0]\\\\=\frac{A_1^2a^2}{2} -(\frac{a}{2}) (0-1)](/tpl/images/0683/0160/779f5.png)
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Having integrated with respect to ϕ and θ, you now have the constant 4π in front of the integral and...
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