Question
Probability of finding the electron $\psi^{2}$ of s orbital doesn't depend upon(a) azimuthal quantum number.(b) energy of s orbital(c) principal quantum number(d) distance from nucleus (r)
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Step 1: The probability of finding an electron in the s orbital is given by the square of the wave function, $\psi^{2}$. Show more…
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The first orbital of $\mathrm{H}$ is represented by: $\psi=\frac{1}{\sqrt{\pi}}\left(1 / a_{0}\right)^{3 / 2} \mathrm{e}^{-t / 2}$ Here, $a_{0}$ is Bohr's radius. The probability of finding the electron at a distance $\mathrm{r}$, from the nucleus in the region $\mathrm{dV}$ is: a. $\psi^{2} 4 \pi r^{2} d r$ b. $\int \psi \mathrm{dV}$ c. $\psi^{2} \mathrm{dr}$ d. $\int \psi^{2} 4 \pi \mathrm{r}^{2} \mathrm{dV}$
For an electron in a hydrogen atom, the wave function is given by $\psi_{1 \mathrm{~s}}$ $=(\pi / \sqrt{2}) e^{-r / a_{0}}$, where $a_{0}$ is the radius of first Bohr's orbit and $r$ is the distance from the nucleus with which probability of finding electron varies. What will be the ratio of probabilities of finding electrons at the nucleus to first Bohr's orbit $a_{\mathrm{o}} ?$ (a) 0 (b) $e$ (c) $e^{2}$ (d) $\frac{1}{e^{2}}$
Atomic Structure
Exercises II
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