Question

The wavefunction for Be\(^{3+}\) in a certain state is given by \(\psi = N e^{-\left(\tfrac{4}{a_0}\right)r}\), where \(N\) is the normalization constant, \(r\) is the distance of the 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

• A. \(4a_0\)
• B. \(\tfrac{a_0}{4}\)
• C. \(8a_0\)
• D. \(\tfrac{a_0}{8}\)

Hint:

For hydrogen-like ions, the 1s most probable radius is given by \(\tfrac{a_0}{Z}\). For Be\(^{3+}\) (\(Z=4\)), the most probable distance is \(\tfrac{a_0}{4}\).

Answer 1

The Correct Option is B

Step 1: Identify the type of wavefunction. 
The given wavefunction is \[ \psi(r) = N e^{-\frac{4r}{a_0}}. \] This is of the form \(\psi(r) \propto e^{-Zr/a_0}\), which corresponds to a hydrogenic 1s orbital with nuclear charge \(Z=4\) (for Be\(^{3+}\)). 
Step 2: Radial probability distribution. 
The probability of finding the electron at a distance \(r\) is given by: \[ P(r) = 4\pi r^2 |\psi(r)|^2. \] So, \[ P(r) \propto r^2 e^{-\frac{8r}{a_0}}. \] 
Step 3: Find most probable radius. 
To maximize \(P(r)\), set derivative to zero: \[ \frac{d}{dr} \left(r^2 e^{-\frac{8r}{a_0}}\right) = 0. \] \[ ⇒ 2r e^{-\frac{8r}{a_0}} - \frac{8}{a_0} r^2 e^{-\frac{8r}{a_0}} = 0. \] \[ ⇒ 2r - \frac{8}{a_0} r^2 = 0. \] \[ ⇒ r \left(2 - \frac{8r}{a_0}\right) = 0. \] Ignoring \(r=0\), \[ r = \frac{a_0}{4}. \] Step 4: Conclusion. Thus, the most probable distance of the electron from the nucleus in this state is \(\tfrac{a_0}{4}\).