Question

The first orbital of H is represented by: $\Psi =\frac{1}{\sqrt{\pi }}{\left[\frac{1}{a_0}\right]}^{3/2}{e}^{-r/a_0}$, where $a_0$ is Bohr orbit. The probability of finding the electron at a distance 'r' from the nucleus is:

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

Answer 1

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

As probability density is ${\Psi }^2$, to calculate probability at a distance of r from nucleus, consider a shell of thickness dr at a distance of r, volume of shell is $4\pi r^2\cdot dr$. So, total probability is ${\Psi }^2\cdot 4\pi r^2\cdot dr$.