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 dV is?

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

Answer 1

The correct option is C ${\Psi }^{2}4\pi r^{2}dr$

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 $={\psi }^{2}4\pi r^{2}dr$.