Answer to Question #190941 in Atomic and Nuclear Physics for Aishwarya Varshney

Question

A research project you are working on requires that an electron have a wave function represented by

𝜓(x) = Ae^−bx for x ≥ 0Ae^+bx for x < 0

The experiment requires that the electron have a 98.0% chance of being within a distance of a = 14.5 cm

 from x = 0,

 on either side of this center point. From this information, you need to determine numerical values for b (in m⁻¹) and the normalization constant A (in m^−1/2) for the wave function.

Answer 1

The probability of finding electron from x = 0 to x = 14.5 cm $= 14.5 \times 10^{-2} \;m$ is P = 98 %

The probability of finding electron:

$P = \int_{x=0}^{0.145} \psi^*(x) \psi(x)dx = 98 \; \% \\ P = \int_{x=0}^{0.145}(\sqrt{b}e^{-bx})^* \sqrt{b}e^{-bx}dx = 98 \; \% \\ \int_{x=0}^{0.145} be^{-2bx}dx = 98 \; \% \\ [\frac{b \times e^{-2bx}}{-2b}]^0.145_0 = 98 \; \% \\ [ e^{-2bx}]^0.145_0 = \frac{98}{100} \times (-2) \\ [ e^{-2b \times 0.145 – e^{-0}}] = \frac{-98}{50} \\ e^{-0.29b} -1 = \frac{-98}{50} \\ e^{-0.29b} = 1 -\frac{98}{50} \\ e^{-0.29b} = \frac{50 -98}{50} \\ e^{-0.29b} = \frac{-48}{50} \\ e^{0.29b} = \frac{-50}{48} = -1.0416 = e^{0.041} \\ b = \frac{0.041}{0.29} = 0.141 \;m^{-1} \\ A = \sqrt{b} = \sqrt{0.141} = 0.376$