Discussion
The wavefunction for Be3+ in a certain state is given by \(\psi = N\,e^{-\left(\frac{4}{a_{0}}r\right)}\), where N is the normalization constant, r is the distance of electron from the nucleus and \(a_{0}\) is the Bohr radius. The most probable distance of the electron from the nucleus in this state is
Answer: B
The most probable distance is \(a_{0}/4\).
The wavefunction \(e^{-4r/a_{0}}\) is the 1s function \(e^{-Zr/a_{0}}\) for \(Z = 4\) (Be3+ is hydrogen-like). The radial probability density is
\[ P(r) \propto r^{2}\psi^{2} = r^{2}e^{-8r/a_{0}} \]
Setting the derivative to zero:
\[ \begin{aligned} 2r - \frac{8r^{2}}{a_{0}} &= 0 \\ r &= \frac{a_{0}}{4} \end{aligned} \]
This is the hydrogen result \(a_{0}\) divided by Z. \(a_{0}/8\) uses the exponent of \(\psi^{2}\) as if it were Z, and \(4a_{0}\) and \(8a_{0}\) multiply by Z instead of dividing; a higher nuclear charge pulls the electron closer, not further out.
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