# Electron wavefunction in He atom

The wavelength for the electron in its lowest energy state in the ion He is proportional to $e^(−2r/a0)$ calculate the probabilities of finding the electron inside a region of volume $V=1.0 pm$ located at (a) the nucleus (b) a distance a a0 from the nucleus.

a) $\psi(r) = A*exp(-2r/a0) a0 =53 pm$

$1 =\int_0^{\infty} \psi^2(r)*d r = (-a0*A^2/4)*exp(-4r/a0) |_0^{\infty} =$

$= 0 + A^2*a0/4 = A^2*a0/4$

$A^2 = \sqrt{(4/a0)}$

$A^2*a0 =4$

for $r = 1 pm$

$probability = \int_{0pm}^{1pm} \psi^2(r)*d r = (-A^2*a0/4)*exp(-4r/a0) |_0^{1pm}=$

$= 1*exp(-4/53) + 1 =-0.9273 +1 = 0.0727$

for $r = 1 pm$ centered on $a_0 =53 pm$

$probability = \int _{53pm}^{54pm} \psi^2(r)*d r = (-A^2*a0/4)*exp(-4r/a0)|_{53pm}^{54pm} =$

$=-1*exp(-4*54/53) +1*exp(-4*53/53) = -0.016984 +0.0183 = 0.0013$