Introduction to Quantum Mechanics 2nd Edition

Published by Cambridge University Press
ISBN 10: 1107179866
ISBN 13: 978-1-10717-986-8

Chapter 1 - Further Problems for Chapter 1 - Page 20: 1.10

Answer

a) P(0)=0; P(1) = 2/25; P(2) = 3/25; P(3) = 5/25; P(4) = 3/25 P(5) = 3/25; P(6) = 3/25; P(7) = 1/25; P(8) = 2/25; P(9) = 3/25 b) Average: =4.72 c) σ = 2.474

Work Step by Step

(a) ans:---π = 3.141592653589793238462643......... there is no zero so the probability of getting zero from the above 25 digits is zero. i.e P(0)=0 similarly, P(1) = 2/25; P(2) = 3/25; P(3) = 5/25; P(4) = 3/25; P(5) = 3/25; P(6) = 3/25; P(7) = 1/25; P(8) = 2/25; P(9) = 3/25 (b) ans:---The most probable is 3 since P(3)=5/25 Median: 13 digits are ≤ 4, 12 digits are ≥ 5, so the median is 4. Average: = (1/25) [0*0 + 1*2 + 2*3 + 3*5 + 4*3 + 5*3 + 6*3 + 7*1 + 8 *2 + 9*3] = (1/25)[0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 118/25 = 4.72. = (1/25)[0 + 1^2 *2 + 2^2 *3 + 3^2 *5 + 4^2 *3 + 5^2 *3 + 6^2 * 3 + 7^2 * 1 + 8^2 *2 + 9^2 *3] = 1/25[0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = 710/25 = 28.4. σ^2 = ^2 = 28.4 − (4.722)^2 = 28.4 − 22.2784 = 6.1216 σ =√(6.1216) = 2.474.
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