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Get Full Access to University Physics - 13 Edition - Chapter 8 - Problem 103p
Get Full Access to University Physics - 13 Edition - Chapter 8 - Problem 103p

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# Antineutrino. In beta decay, a nucleus emits an electron.

ISBN: 9780321675460 31

## Solution for problem 103P Chapter 8

University Physics | 13th Edition

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Problem 103P

Antineutrino.? In beta decay, a nucleus emits an electron. A 210Bi (bismuth) nucleus at rest undergoes beta decay to 210Po (polonium). Suppose the emitted electron moves to the right with a momentum of The 210Po nucleus, with mass 3.50 X 10-25 kg, recoils to the left at a speed of 1.14 X 103 m/s. Momentum conservation requires that a second particle, called an antineutrino, must also be emitted. Calculate the magnitude and direction of the momentum of the antineutrino that is emitted in this decay.

Step-by-Step Solution:

Solution 103P Step 1: Momentum of the emitted electron towards right is, P = 5.e0 × 10 22 kg The mass of 210Po nucleus is, m = 3.50 × 10 25 kg . Po Velocity of the nucleus is, V = 1.14 × 10 m/stowards left. Po 25 3 22 The momentum of the nucleus will be, P Po = 3.50 × 10 × 1.14 × 10 = 3.99 × 10 .

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