Question

The complete wave function of an orbital of hydrogen is given as $\psi =\frac{1}{4a_0\sqrt{2\pi a_0}}(2-\frac{r}{a_0})e^{\frac{-r}{2a_0}}$,
($a_0$ is first Bohr radius)
Which of the following option(s) is/are correct?

• A. Orbital may be 3p
• B. Angular node is at $2a_0$ from nucleus
• C. Ratio of density of probability of finding electron at distance $a_0$ to nucleus is $\frac{e^{-1}}{4}$
• D. Orbital has a total of one node

Answer 1

Answer: (C), (D)

The correct options are
C Ratio of density of probability of finding electron at distance $a_0$ to nucleus is $\frac{e^{-1}}{4}$
D Orbital has a total of one node

a) The given wave function has no dependence on any angles $\theta ,\varphi$, it must represent an s orbital.
b) There will be no angular nodes because it is an s orbital so $l=0$.
c) Probability of finding an electron at $a_0$ and nucleus ($r=0$) are respectively
${\psi }_{a_o}^2=\left[\frac{1}{4a_0\sqrt{2\pi a_0}}\right(2-\frac{a_o}{a_0})e^{\frac{-a_o}{2a_0}}{]}^2$
${\psi }_o^2=\left[\frac{1}{4a_0\sqrt{2\pi a_0}}\right(2-\frac{0}{a_0})e^{\frac{0}{2a_0}}{]}^2$
So $\frac{{\psi }_{a_o}^2}{{\psi }_o^2}=\frac{e^{-1}}{4}$
d) Since the orbital wavefunction will be zero for only $r=2a_o$, hence total number of nodes will be 1.