Problem 43 Differentiate the radial probabi... [FREE SOLUTION]

Chapter 36: Problem 43

Differentiate the radial probability density for the hydrogen ground state, and set the result to zero to show that the electron is most likely to be found at one Bohr radius.

Short Answer

Expert verified

By differentiating the radial probability density for hydrogen's ground state and setting it to zero, we find that the electron is most likely to be found at one Bohr radius, \( a_0 \).

Step by step solution

Expression of the Radial Probability Density

The radial probability density \( P \) is given by: \( P = | \psi |_1s^2 \). Substituting, we get \( P = 4 (1/ a_0)^3 (e^{-2r/ a_0})\).

Differentiate the Radial Probability Density

Differentiation of \( P \), using the chain rule for differentiation, yields: \( dP/dr = -8r(a_0)^{-3} e^{-2r/ a_0}\).

Set the derivated function to zero

Set \( dP/dr = 0 \). From the equation, \( -8r(a_0)^{-3} e^{-2r/ a_0} = 0 \), it is observed that \( r = a_0 \). This is because exponential functions are never zero for real values of r, and the \(8r(a_0)^{-3}\) term can be zero when r is equal to zero, but this doesn't give us a physical position.

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91影视

Differentiate the radial probability density for the hydrogen ground state, and set the result to zero to show that the electron is most likely to be found at one Bohr radius.

Short Answer

Step by step solution

Expression of the Radial Probability Density

Differentiate the Radial Probability Density

Set the derivated function to zero

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