91Ó°ÊÓ

To be consider an electron within in the ground state of a hydrogen atom

The region in an atom where the probability of finding an electron is 90% is known as orbital.

The following figure shows plots of E and U(r)

b)

At ground state, the electron has energy.

$$\begin{gathered} E_1=-\frac{m{e}^4}{2{\left(4\mathrm{π}{∈}_0\right)}^2{n}^2} \\ =\left(-\frac{e^2}{4\mathrm{π}{∈}_0}\right)\frac{1}{2a_0} \end{gathered}$$

Where, m represents mass of electron

$\frac{1}{4\mathrm{π}{∈}_0}$represents Coulomb constant and$a_0$represents Bohr radius

Potential energy is given by

P E=$-\frac{e^2}{4\mathrm{π}{∈}_{0}r}$

The farthest an electron can get is at the location where the total energy equals potential energy

$E_1=PE$

We know, Both energies are equal at radius of$2a_0$

Distance of electron from proton is r = $2a_0$.

c)

The radius for ground state is given by

$R_{1,0}\left(r\right)=\frac{1}{{\left(a_0\right)}^{3/2}}2e^{-r/a_0}$

The probability density is given by

$P\left(r\right)=r^2{\mathrm{R}}_{1.0}^2\left(r\right)$

Substitute$\frac{1}{{\left(a_0\right)}^{3/2}}2e^{-2r/a_0}$for$R_{1,0}\left(r\right)$

P(r) =$\frac{4}{a_0^3}{r}^2{e^{-2r/a}}_0$

The probability to find the electron at classical forbidden region is

$P={∫}_{2a_0}^{∞}\frac{4}{a_0^3}{r}^2{e}^{-2r/a_0}dr$

Substituting x for localid="1659327226770" $\frac{2r}{a_0}$, we get

localid="1659327221963" $$\begin{gathered} P=\frac{1}{2}{∫}_4^{∞}{x}^2{e}^{-x}dx \\ =-\frac{1}{2}\left({\overline{)x^2{e}^{-x}}}_4^{∞}-{∫}_4^{∞}2x{e}^{-x}dx\right) \\ =8e^{-4}+{∫}_4^{∞}x{e}^{-x}dx \\ =8e^{-4}\left({\overline{)x{e}^{-x}}}_4^{∞}-{∫}_4^{∞}x{e}^{-x}dx\right) \end{gathered}$$

On solving further

localid="1659327210389" $\begin{array}{l}P=8e^{-4}+4e^{-4}+{∫}_4^{∞}{e}^{-x}dx\\ =13e^{-4}\\ =0.238\end{array}$

The probability to find the particle at classical forbidden region is 0.238