The first orbital of H is represented by
$ψ = \frac{1}{\sqrt{π}} (\frac{1}{a_{0}}) e^{- r / a_{0}}$ , 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

Question

The first orbital of H is represented by ψ = 1 π 1 a 0 e - r / 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

TestHub · Accepted Answer

This can be obtained by the solution of Schrodinger wave equation as follows
Probability $= ψ^{2} d V$
As orbital is spherically symmetrical
So $V = \frac{4}{3} π r^{3}$
i.e., $\frac{d V}{d r} = 4 π r^{2}$
Hence, probability $= ψ^{2} 4 π r^{2} d r$ .

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