Question

The first orbital of $H$ is represented by $\psi=\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2} e^{-\pi / a_{0}}$, where $a_{0}$ is Bohr's radius. The probability of finding the electron at a distance $r$ from the nucleus in the region $d V$ is

• A. $\psi^{2}dr $
• B. $\int \Psi^{2} 4 \pi r^{2} d V$
• C. $\Psi^{2} 4 \pi r ^{2} dr$
• D. $\int \Psi dV$

Answer 1

Answer: (C)

This is obtained by the solution of Schrodinger wave equation.
Probability $=\psi^{2} dV$
Its orbital is spherically symmetrical. Thus,
volume, $V =\frac{4}{3} \pi r ^{3}$
$\therefore \frac{ dV }{ dr }=4 \pi r ^{2}$
$\because$ Probability $=\psi^{2} 4 \pi r^{2} dr .$