Start with the expression for the Heisenberg uncertaintyprinciple and rearrange

Question

Start with the expression for the Heisenberg uncertaintyprinciple and rearrange the units to obtain a relationshipbetween uncertainty in energy and uncertainty in time. a) How much time is required to measure the energy of a 1keVelectron with an accuracy of 99%? b) If an unstable particle exists for a microsecond beforedecaying into a different particle, when you try to measureits energy, what is the minimum uncertainty in your result? c) What if the decay process took a femtosecond instead? d) An attosecond?
I promise to rateyou as lifesaver if you SHOW ALL WORK, with calculations andunits! Start with the expression for the Heisenberg uncertaintyprinciple and rearrange the units to obtain a relationshipbetween uncertainty in energy and uncertainty in time. a) How much time is required to measure the energy of a 1keVelectron with an accuracy of 99%? b) If an unstable particle exists for a microsecond beforedecaying into a different particle, when you try to measureits energy, what is the minimum uncertainty in your result? c) What if the decay process took a femtosecond instead? d) An attosecond?
I promise to rateyou as lifesaver if you SHOW ALL WORK, with calculations andunits!

Explanation / Answer

The Heisenberg uncertainty principle is x * p = (h/2) ---------(1) x is the uncertainty in the measurement ofposition,p is the uncertainty in the measurement ofmomentum and h is planck's constant and has a value 6.63 *10-34 J-s Dividing equation (1) by t we get (x/t) * p = (h/2) * (1/t) or v * m * v = (h/2) * (1/t) or m * v2 = (h/2) * (1/t)--------(2) the kinetic energy of the electron is K = (1/2)m * v2 or m * v2 = 2K ---------(3) from (2) and (3) we get 2K = (h/2) * (1/t) or t = (2/h) * 2K ---------(4) K = E - (99E/100) = (E/100) where E = 1 keV = 1 *103 * 1.6 * 10-19 J = 1.6 * 10-16J or K = (1.6 * 10-16/100) = 1.6 *10-18 J b)we know from equation (4) that t = (2/h) * 2K or K = (h/2) * t * (1/2) = (1/4) * t---------(5) t = 1 * 10-6 s or K = (1/4 * 3.14) * 1 * 10-6 = 0.0769 *10-6 therefore the minimum uncertainty in the measurement of energyis 0.0769 * 10-6. c)when the decay process took t = 1 * 10-15 s thensimilarly the uncertainty in the measurement of energy of theustable particle is K = 0.0769 * 10-15 d)when the decay process took t = 1 * 10-18 s thensimilarly the uncertainty in the measurement of energy of theustable particle is K = 0.0769 * 10-18 d)when the decay process took t = 1 * 10-18 s thensimilarly the uncertainty in the measurement of energy of theustable particle is K = 0.0769 * 10-18 K = 0.0769 * 10-18

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