Question

The maximum radial probability in 1s-orbital occurs at a distance when : [ro = Bohr radius ]

• A. r = ro
• B. r = 2ro
• C. r = \frac{r_0}{2}
• D. 2r = \frac{r_0}{2}

Answer 1

Answer: (A)

r = ro

Answer 2

The radial distribution function for a 1s orbital is $P(r) = \frac{4}{r_0^3} r^2 e^{-2r/r_0}$. To find the maximum, we set the derivative $\frac{dP(r)}{dr}$ to zero. $\frac{dP(r)}{dr} = \frac{4}{r_0^3} e^{-2r/r_0} \left( 2r - \frac{2r^2}{r_0} \right)$. Setting this to zero and considering the non-zero solution gives $2r(1 - \frac{r}{r_0}) = 0$, which yields $r = r_0$.