Answered: (a) The lifetime of a highly unstable… | bartleby

Related questions

Question

PRACTICE ANOTHER
MY NOTES
ASK YOUR TEACHER
(a) The lifetime of a highly unstable nucleus is 10-13 s. What is the smallest uncertainty (in ev) in its decay energy?
eV
(b) What is the ratio of this energy, AE, to the rest energy of an electron, Erest?
AE
Erest
Additional Materials
O Reading

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 9 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

A certain medium has the following properties: Σs = 0.1 cm−1,Σa = 0.01 cm−1. a) What is the probability that a neutron will travel 1 cm in this medium without interacting with a nucleus? b) If the neutron speed is 2 ×105 cm s−1, what is the average time between scattering collisions?
A neuto is moving at a speed of 71 m/s a.) The debregoli wavelength associated with the neutron is _________ nm. b.) If another neutron B travels at four times the speed of neutron A, the ratio of the wavelengths of neutron B to neutron A will be ______. c.) If the uncrtainty of measuring the speed of neutrons is 2.8 m/s The min. uncertainty in measuring the position of the neutron is _____m.
Now the energy of a neutrino is measured at a specific time. If the uncertainty in the time measurement is 2.31 x 10-24 seconds, what is the minimum uncertainty in the energy?
A proton or a neutron can sometimes "violate" conser- vation of energy by emitting and then reabsorbing a pi meson, which has a mass of 185 MeV/c². This is possi- ble as long as the pi meson is reabsorbed within a short enough time At consistent with the uncertainty principle, (a) Consider p → p + л. By what amount AE is energy conservation violated? (Ignore any kinetic energies.) (b) For how long a time At can the pi meson exist? (c) Assuming that the pi meson travels at very nearly the speed of light, how far from the proton can it go? (This procedure, as we discuss in Chapter 12, gives us an
Given: Nc = (2.51x1019)(mn/mo)3/2 (T/300)3/2 Nv = (2.51x1019)(mp/mo)3/2 (T/300)3/2 and ni = (NcNv)1/2 e(-Eg/2kT) show that: ni = (2.51x1019) ((mn/mo) * (mp/mo) )3/2 e(-Eg/2kT)
Compare the gravitational potential energy, mgh, exerted on an electron 100 m above theearths surface to the potential energy exerted on an electron by a hydrogen nucleus at adistance of 1 Angstrom. What does this imply?