Final answer:
The smallest uncertainty in the decay energy of a highly unstable nucleus is approximately 6.63 x 10^(-34) J. This uncertainty is much smaller than the rest energy of an electron, which is approximately 9.31 x 10^(-31) J.
Step-by-step explanation:
The smallest uncertainty in the decay energy of a highly unstable nucleus can be calculated using the uncertainty principle of energy and time. According to the Uncertainty Principle, the product of the uncertainty in energy (ΔE) and the uncertainty in time (Δt) is equal to or greater than ħ/2.
Given that the lifetime (Δt) of the nucleus is 10^(-20) seconds, we can substitute this value into the equation to find the smallest uncertainty in its decay energy. Using the equation ΔE * Δt >= ħ/2, we find that the smallest uncertainty in the decay energy is approximately 6.63 x 10^(-34) J, which is option a) in the given choices.
To compare this uncertainty with the rest energy of an electron, we can use Einstein's equation E = mc^2, where E is energy, m is mass, and c is the speed of light. The rest energy of an electron can be calculated by multiplying its mass by the speed of light squared.
The rest energy of an electron is approximately 9.31 x 10^(-31) J. The uncertainty in the decay energy of the highly unstable nucleus (6.63 x 10^(-34) J) is much smaller than the rest energy of an electron (9.31 x 10^(-31) J).