Atomic Structure Question 424
Question: Let $ m_{p} $ be the mass of a proton, $ m_{n} $ that of a neutron, $ M_1 $ that of a $ _10^{20}Ne $ nucleus and $ M_2 $ that of a $ _20^{40}Ca $ nucleus. Then
Options:
A) $ M_2=2M_1 $
B) $ M_1<10(m_{p}+m_{n}) $
C) $ M_2>2M_1 $
D) $ M_1=M_2 $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ _{20}^{40}Ne $ contains 10 protons and 10 neutrons
$ \therefore M _{1}=10m _{p}+10m _{n} $
$ _{20}^{40}Ca $ contains 20 protons and 20 neutrons
$ \therefore M _{2}=20m _{p}+20m _{n} $
$ \therefore M _{2}=2M _{1} $
