Question and Answer - Nuclear Physics #15 | Princeton University - Edubirdie

University:
Princeton University
Course:
PHY 103 | General Physics I
Academic year:

2023
Views:

345

Pages:

1
Author:

abdulsalam23745

eyes.17APage-26ifz 2 then 2= 2+01A13D15.8A - -18.3A2/3 - 0.714 Z(z-1) 23.2 (A-22)2205/mmA 1/3ADB =0=- 0.714 (22-1)azA 1/3+ 23.2x4 (A-22)=01.29A- -0.714F 2/3 (2z-1) + 9.2.8 (A-22)OurA-0.714 A x2Z+0.714A. 2/3 + 92.8A - 185.6Z =02/30.714A2/3 + 92.81 AZ =185.6 + 0.714A2/3 X2Z= 28.547Z28.547 = 0.446 1 0.45=A64(option-9

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