Because the factor h on the right side of the Heisenberg uncertainty principle has units of Joule-seconds,it suggests that the energy of a system also has uncertainty.The uncertainty in energy depends on the length of the time interval during which a system exists.$\Delta$E$\Delta$t $\ge$ h/4 $\pi$.Suppose an unstable mass is produced during a high-energy collision such that the uncertainty in its mass is m_e/100.(m_e = 9.11 *10 ^- ³¹ kg. )How long will this particle exist? A)6.4 * 10 ^- ²⁰ s B)2.3 *10 ^- ²³ s C)1.0 * 10¹⁵ s D)1.2 * 10¹³ s E)8.1 * 10 ^- ¹⁹ s

Question

Quizplus · Accepted Answer

The uncertainty in energy is given by the Heisenberg uncertainty principle:$$\Delta E \Delta t \geq \frac{h}{4\pi}$$where $\Delta E$ is the uncertainty in energy, $\Delta t$ is the uncertainty in time, and $h$ is Planck's constant.We are given that the uncertainty in mass is $m_e/100$, where $m_e$ is the mass of an electron. We can use the mass-energy equivalence equation to convert this to an uncertainty in energy:$$\Delta E = (\Delta m)c^2$$where $c$ is the speed of light.Substituting this into the Heisenberg uncertainty principle, we get:$$(\Delta m)c^2 \Delta t \geq \frac{h}{4\pi}$$Solving for $\Delta t$, we get:$$\Delta t \geq \frac{h}{4\pi (\Delta m)c^2}$$Substituting the given values, we get:$$\Delta t \geq \frac{6.63 	imes 10^{-34} 	ext{ J s}}{(4\pi)(9.11 	imes 10^{-31} 	ext{ kg})(3 	imes 10^8 	ext{ m/s})^2}$$$$\Delta t \geq 6.4 	imes 10^{-20} 	ext{ s}$$Therefore, the particle will exist for at least $6.4 	imes 10^{-20}$ seconds.

Because the Factor H on the Right Side of the Heisenberg