Question

An electron in an atom can "jump down" from a higher energy level to a lower one, then to a lower one still. The energy the atom thus loses at each jump goes to a photon. Typically, an electron might occupy a level for a nanosecond. What uncertainty in the electron's energy does this imply?

Short answer

The uncertainty in the electron’s energy is $∆E\ge 5.3\times {10}^{-23}J$.

Step-by-step solution

Given data

Time is given as: $∆t={10}^{-9}s$.

Uncertainty principle

The principle states that the position and the velocity of an object cannot be measured with 100 % accuracy at the same time.

$∆x·∆p\ge \frac{\sout{h}}{2}$

$∆x$= Uncertainty in the position.

$∆p$ = Uncertainty of momentum.

$\sout{h}$ = Planck's constant. = 1.05x10^-34 J.s

Electron’s energy

The Energy and Time Uncertainty Principle,

$∆t·∆E\ge \frac{\sout{h}}{2}$

Substituting values, and we get:

role="math" localid="1658386938701" $$\begin{gathered} ∆E\ge \frac{1.05\times {10}^{-34}}{2\times 1\times {10}^{-9}} \\ ∆E\ge 5.3\times {10}^{-26}J \end{gathered}$$

Therefore, the total energy is $∆E\ge 5.3\times {10}^{-26}J$.