Question

Please answer the following problem completely, answer both parts (A and B). Show your work, write LEGIBLY, and label the solutions. Thank you

In the CHOOZ neutrino oscillation experiment, from a nuclear reactor were studied as a function of the distance to the reactor core. At a distance of 250 m, the observed event rate was 0.95±0.10 of the expected rate. At a mean anti-neutrino energy of 5 MeV, what limit does this place on a possible neutrino mass difference, assuming maximum mixing (0p 45°)?

Answer 1

In the CHOOZ neutrino oscillation experiment V_c^bar were distance at 250 m In this problem observed went rate is 0.95 plusminus 0.10 This is the survival probability of the anti-neutrino conving from the reactor. The oscillation probability and survival probability can be related as p_oscillation = 1 - p_survival for simplicity p_curival = 0.95 [ignoring the basic] p_oscillation = 1 - 0.95 = 0.05 we know that, E or two neutrino to oscillate their oscillate probability is written in terms of thus mass square difference, energy and length, therefore, oscillation = sin^2 2 theta, sin^2 [1.27 dekta m^2 L/E v^bar] equation (1) \r\n where delta m^2 is in cv^2, L is in km, E-v^bar is in Ger E_v^bar = 5 Mev = 5 times 10^-3 Gev L = 0.25 km op = 45 degrees p_oscillation = sin^2 pi/2 sin^2 [1.27 times delta m^2 times 0.15)/5 times 10^-3] = 0.05 delta m^2 = 5 times 10^-3/1.27 times 0.25 sin^-1 [squareroot 0.05