Answer to Question #248766 in Physics for KUMARESAN K

Question

The wave function of the hydrogen atom in its ground state (1s) is ψ=(πa03)−12exp(−ra0 where a0=0.529×10−8 cm. The charge density is ρ(x,y,z)=−e|ψ|2, according to the statistical interpretation of the wave function.
Then the value of ⟨r⟩ is given by

a, 5a0^2

b, 2a0^2

c, 4a0^2

d, 3a0^2

Answer 1

The wave function

$$\psi(r)=\frac{1}{\sqrt{\pi}a_0^{3/2}}\exp(-r/a_0)$$

The average value of $r^2$ is given by

$$\langle r^2\rangle=4\pi \int_0^{\infty}r^2\psi^2r^2dr=\frac{4}{a_0^3}\int_0^{\infty}\exp(-2r/a_0)r^4dr$$

$$=\frac{4}{a_0^3}*\frac{3a_0^5}{4}=3a_0^2$$

Answer: d