A) 24.9 amu
B) 34.9 amu
C) 54.9 amu
D) 35.289 amu
Correct Answer: B
Solution :
[b] \[m=17{{m}_{p}}+18{{m}_{n}}\] \[=17\times 1.007825+18\times 1.008665\] = 35.289 \[\Delta m=\frac{298MeV}{931.2MeV/amu}=0.3200\,amu\] \[\therefore \text{ }Atomic\text{ }mass=m-\Delta m\] \[=35.289-0.32\] = 34.969 amuYou need to login to perform this action.
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