6. Calculate the most probable radius at which an electron will be found when it occupies a 1s orbital of a hydrogen atom.
$$
\text { hint: } \begin{aligned}
& \Psi_{1 s}(x)=\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_{0}}\right)^{\frac{3}{2}} e^{-\frac{x}{a_{0}}} \text {; } a_{\circ} \text { is the Bohr radius } \
& \text { radial probabilit } p(r)=4 \pi r^{2}\left|\Psi_{1 s}(x)\right|^{2}
\end{aligned}
$$

Question

6. Calculate the most probable radius at which an electron will be found when it occupies a 1s orbital of a hydrogen atom.
$$
\text { hint: } \begin{aligned}
& \Psi_{1 s}(x)=\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_{0}}\right)^{\frac{3}{2}} e^{-\frac{x}{a_{0}}} \text {; } a_{\circ} \text { is the Bohr radius } \
& \text { radial probabilit } p(r)=4 \pi r^{2}\left|\Psi_{1 s}(x)\right|^{2}
\end{aligned}
$$

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Express the radial probability density function**: The radial probability density function $p(r)$ for an electron in a 1s orbital is given by the square of the radial part of the wave function $|\Psi_{1s}(r)|^2$ multiplied by the surface area of a sphere $4\pi r^2$. ‖ ‖ 2 ⋮ **Substitute the wave function**: The given wave function for the 1s orbital is $\Psi_{1s}(r) = \frac{1}{\sqrt{\pi}}\left(\frac{1}{a_0}\right)^{\frac{3}{2}} e^{-\frac{r}{a_0}}$. We need to square this and multiply by $4\pi r^2$ to get $p(r)$. ‖ ‖ 3 ⋮ **Calculate $p(r)$**: The radial probability density function is $p(r) = 4\pi r^2 \left|\Psi_{1s}(r)\right|^2 = 4\pi r^2 \left(\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_0}\right)^{\frac{3}{2}} e^{-\frac{r}{a_0}}\right)^2$. Simplifying this gives $p(r) = \frac{4r^2}{a_0^3} e^{-\frac{2r}{a_0}}$. ‖ ‖ 4 ⋮ **Find the maximum of $p(r)$**: To find the most probable radius, we need to find the value of $r$ that maximizes $p(r)$. This is done by taking the derivative of $p(r)$ with respect to $r$ and setting it equal to zero. ‖ ‖ 5 ⋮ **Differentiate and solve for $r$**: Differentiating $p(r)$ with respect to $r$ gives $\frac{dp(r)}{dr} = \frac{4}{a_0^3}(2r e^{-\frac{2r}{a_0}} - \frac{2r^2}{a_0}e^{-\frac{2r}{a_0}})$. Setting $\frac{dp(r)}{dr} = 0$ and solving for $r$ gives $r = a_0$. ‖ ∻Answer∻ ⚹ The most probable radius at which an electron will be found when it occupies a 1s orbital of a hydrogen atom is the Bohr radius $a_0$. ⚹ ∻Key Concept∻ ⚹ The most probable radius for an electron in a 1s orbital is found by maximizing the radial probability density function. ⚹ ∻Explanation∻ ⚹ The most probable radius is the value of $r$ at which the radial probability density function $p(r)$ reaches its maximum. For a 1s orbital, this occurs at the Bohr radius $a_0$. ⚹◊What is the significance of the radial probability density function when determining the most probable radius of an electron in a hydrogen atom's 1s orbital?⍭ Generate me a similar question◊

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