Question

Assertion :All $s$-orbital in $H$-atom corresponds to a non-zero probability density at nucleus. Reason: The probability density is given by $:{\psi }^2$ and $\psi \propto e^{\frac{-Zr}{2a_0}}$.

• A. Both Statement and Explanation are correct and Explanation is the correct explanation for Statement.
• B. Both Statement and Explanation are correct but Explanation is not the correct explanation for Statement.
• C. Statement is correct but Explanation is wrong.
• D. Statement is wrong but Explanation is correct.

Answer 1

The correct option is A Both Statement and Explanation are correct and Explanation is the correct explanation for Statement.

The square of the wave function $\left({\psi }^2\right)$ gives the probability density.
Since, $\psi \propto e^{\frac{-Zr}{2a_0}}$
${\psi }^2\propto (e^{\frac{-Zr}{2a_0}}{)}^2$
At the nucleus, $r=0$
Hence, ${\psi }^2\propto e^{-0}\propto 1$.
Thus, an s-orbital in H-atom corresponds to a non-zero probability density at nucleus.
Both Statement and Explanation are correct and Explanation is the correct explanation for Statement.