Question

Problem 10 (Problem 2.24 in textbook) The wavefunction for the electron in a hydrogen atom in its ground state (the 1s state for which n 0, l-0, and m-0) is spherically symmetric as shown in Fig. 2.14. For this state the wavefunction is real and is given by exp-r/ao h2Eo 5.29 x 10-11 m. This quantity is the radius of the first Bohr orbit for hydrogen (see next chapter). Because of the spherical symmetry of ịpo, dV in Eq. (2.56) is dV- 4πγ2dr and the integral in Eq. (2.56) can be written as Where a2/ao. (a) Verify that the required normalization required by (2.56) is satisfied, i.e., the electron is somewhere in the space around the proton. (b) What is the probability the electron is found a radial distance r<a from the proton?

Answer 1

a) For the wavefunction to be normalized

$\int_{0}^{\infty}\psi_{0}(r)\psi^{*}_{o}(r) dV = 1$ So now we calclate this integral using the function given

$\\ I = \int_{0}^{\infty}\psi_{0}(r)\psi^{*}_{o}(r) dV \\ =\int_{0}^{\infty} \frac{1}{\sqrt{\pi a_{o}^{3}}}e^{-r/a_{o}}\frac{1}{\sqrt{\pi a_{o}^{3}}}e^{-r/a_{o}} 4\pi r^2 dr \\ =\frac{4}{a_{o}^{3}}\int_{0}^{\infty} r^2 e^{-2r/a_{0}} dr$

Now to solve the definite integral I₁ of I we have to employ integration by parts which is

$\int uv dx = u\int vdx - \int \left [\frac{du}{dx}\int vdx \right]dx$

Putting u= r² and v= exp(-2r/a_o), we get

$I_{1} = \frac{4}{a_{o}^{3}} \int r^2 e^{-2r/a_{0}} dr \\ =\frac{4}{a_{o}^{3}} \left[ r^2 \int e^{-2r/a_{o}}dr - \int \left(2r\int e^{-2r/a_{0}}dr \right )dr \right ] \\ =\frac{4}{a_{o}^{3}} \left[r^{2} \frac{e^{-2r/a_{0}}}{-2r/a_{o}} - \int \left (2r \frac{e^{-2r/a_{0}}}{-2r/a_{o}} \right )dr \right ]$

Now putting u = r and v=exp(-2r/a_o), we get

$I_{1} = \frac{4}{a_{o}^{3}} \left[r^{2} \frac{e^{-2r/a_{0}}}{-2r/a_{o}} - \int \left (2r \frac{e^{-2r/a_{0}}}{-2r/a_{o}} \right )dr \right ] \\ = \frac{4}{a_{o}^{3}} \left[\frac{-r e^{-2r/a_{o}}}{2} + a_{o} \int re^{-2r/a_{o}} dr \right ] \\ = \frac{4}{a_{o}^{3}}\left [ \frac{-r e^{-2r/a_{o}}}{2} + a_{o} \left (r \frac{e^{-2r/a_{o}}}{-2r/a_{o}} - \int \frac{e^{-2r/a_{o}}}{-2r/a_{o}} dr\right )\right ] \\$

Solveing this we get

$I_{1} = \frac{4}{a_{o}^3} e^{-2r/a_{o}} \left [ -\frac{a_{o}^{3}}{4} - \frac{a_{o}^{2}r}{2} -\frac{a_{o}r^{2}}{2} \right ] + C$

Putting the limits 0 to infinity we get

I = 1

Therefore the wavefunction is normalized

To find the probability of finding the electron within a_o we put the limits from 0 to a_o

then we get probability (P)

$P = 1-\frac{5}{e^2}$