Answer:
Step-by-step explanation:
[tex]\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)[/tex]
[tex]=A_1^2\int\limits^{\infty}_0 r^2e^{-2r/a}\ dr[/tex]
[tex]=A_1^2[\frac{r^2e^{2r/a}}{-2/a} |_0^{\infty}-\int\limits^{\infty}_0 2r\frac{e^{-2r/a}}{-2/a} \ dr][/tex]
[tex]=A^2_1[0+\int\limits^{\infty}_0 a\ r\ e^{-2r/a}\ dr][/tex]
[tex]=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][/tex]
[tex]=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[/tex]
[tex]=\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)[/tex]
[tex]=\frac{A_1^2a^3}{4}[/tex]
[tex]\therefore A_1^2\int\limits^{\infty}_0 r^2(e^{-r/a}) \ dr =\frac{A_1^2a^3}{4}[/tex]
Find the unique positive value of A1
[tex]=4\pi (\frac{A_1^2a^3}{4} )\\\\=A_1^2a^3\pi\\\\A_1^2=\frac{1}{a^3\pi} \\\\A_1=\sqrt{\frac{1}{a^3\pi} }[/tex]
