Question

Consider an electron in the energy eigenstate \(\psi_{211}(\vec{r})\) of the hydrogen atom. Given that the radial probability distribution of the electron in such a state takes its maximum value at \(r = n_0 a\), where \(a\) is the Bohr radius, and \(n_0\) is an integer. The value of \(n_0\) (in integer) is . The radial part of the wavefunction \(\psi_{211}(\vec{r})\) is given by \(R_{21}(r) = \dfrac{1}{\sqrt{24a^{5}}}\,re^{-r/2a}\).

Hint:

Hint:
\(P(r)=r^2R_{21}^2(r)\propto r^4e^{-r/a}\); maximize it by setting its derivative to zero, or by maximizing \(\ln[r^4e^{-r/a}]\).

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
The radial probability distribution tells us how likely we are to find the electron at a distance \(r\) from the nucleus, regardless of direction. For a state with radial wavefunction \(R_{nl}(r)\), this distribution is \(P(r) = r^2 |R_{nl}(r)|^2\); the extra \(r^2\) comes from the volume of a thin spherical shell of radius \(r\). We are given \(R_{21}(r) = \dfrac{1}{\sqrt{24a^5}}\,re^{-r/2a}\), and need the value of \(r\) where \(P(r)\) is maximum.

Step 2: Key Formula or Approach:
\[ P(r) = r^2\big[R_{21}(r)\big]^2 = r^2\cdot\dfrac{1}{24a^5}\,r^2e^{-r/a} = \dfrac{r^4e^{-r/a}}{24a^5} \]
To maximize \(P(r)\), set \(\dfrac{dP}{dr}=0\) and solve for \(r\), keeping the physically meaningful maximum and ignoring the trivial \(r=0\) minimum.

Step 3: Detailed Explanation:
Differentiate \(r^4e^{-r/a}\) using the product rule:
\[ \dfrac{d}{dr}\big(r^4e^{-r/a}\big) = 4r^3e^{-r/a} + r^4\left(-\dfrac{1}{a}\right)e^{-r/a} = r^3e^{-r/a}\left(4 - \dfrac{r}{a}\right) \]
Setting this to zero: since \(e^{-r/a}\) is never zero, either \(r^3=0\) (giving \(r=0\), where the probability is zero, not a maximum) or:
\[ 4 - \dfrac{r}{a} = 0 \implies r = 4a \]
This is the physical maximum, since \(P(r)\) rises from \(0\) at \(r=0\), reaches this peak, then decays back to \(0\) as \(r\to\infty\).

Step 4: Final Answer:
Comparing \(r=4a\) with \(r=n_0a\), we get \(n_0=4\). \[ \boxed{n_0 = 4} \]