Question

(a) If the position of an electron in a membrane is measured to an accuracy of 1.00 μm , what is the electron’s minimum uncertainty in velocity?

(b) If the electron has this velocity, what is its kinetic energy in eV?

(c) What are the implications of this energy, comparing it to typical molecular binding energies?

Short answer

(a) The electron’s minimum uncertainty in velocity is: 57.9 m/s.

(b) The kinetic energy is obtained as: 9.5 x 10^{- 19} eV.

(c) When the energy required by molecule binding energies is compared, then this energy is substantially lower.

Step-by-step solution

Define Quantum Physics

Quantum physics is the study of matter and energy at their most basic levels. Its objective is to have a better understanding of the traits and actions of the basic constituents of nature.

Evaluating the electron’s minimum uncertainty in velocity

(a)

To obtain the minimum uncertainty in the speed.

The equation used here is:\[\Delta x\Delta p{\rm{ }} = {\rm{ }}\frac{h}{{4\pi }}\].

Here the value is:\[\Delta p{\rm{ }} = {\rm{ }}m\Delta v\].

\[\begin{array}{c}\Delta xm\Delta v = \frac{h}{{4\pi }}\\\Delta v = \frac{h}{{4\pi \Delta xm}}\\ = \frac{{6.63 \times {{10}^{ - 34}}j \bullet s}}{{4\pi \times 1.00 \times {{10}^{ - 6}}m \times 9.11 \times {{10}^{ - 31}}kg}}\\ = 57.9m/s\end{array}\]

Therefore, the electron’s minimum uncertainty in velocity is: \[57.9{\rm{ }}m/s\].

Evaluating the kinetic energy

(b)

To evaluate the kinetic energy in electron volts.

The equation used is:\[K.E{\rm{ }} = {\rm{ }}\frac{{m{v^2}}}{{2e}}\].

Here, the value of \[e\] is said to be the electron charge.

\[\begin{array}{c}K.E{\rm{ }} = {\rm{ }}\frac{{m{v^2}}}{{2e}}\\ = \frac{{9.11 \times {{10}^{ - 31}}kg \times {{(57.9{\rm{ }}m/s)}^2}}}{{2 \times 1.6 \times {{10}^{ - 19}}C}}\\ = 9.5 \times {10^{ - 19}}eV\end{array}\]

Therefore, the kinetic energy is: \[9.5 \times {10^{ - 19}}eV\].

Explaining the implications

(c)

This energy is significantly small as compared to the energy needed with the molecular binding energies.

Therefore,when compared to the energy required by molecule binding energies, this energy is substantially lower.