Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Let nip be the mass of proton,\[{{m}_{p}}\]the mass of neutron.\[{{M}_{1}}\]the mass of\[_{10}^{20}Ne\]nucleus and\[{{M}_{2}}\]the mass of\[_{20}^{40}Ca\]nucleus. Then: \[{{M}_{2}}=2{{M}_{1}}\] \[{{M}_{2}}<2{{M}_{1}}\] \[{{M}_{1}}<10({{m}_{n}}+{{m}_{p}})\] \[{{M}_{2}}<2{{M}_{1}}\]A) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 are correct
D) 1, 2 and 3 are correct
Correct Answer: B
Solution :
Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is always less than the sum of masses of its constituent particles. \[_{10}^{20}Ne\]is made up of 10 protons and 10 neutrons. Therefore, mass of\[_{10}^{20}Ne\]nucleus \[{{M}_{1}}<10({{m}_{p}}+{{m}_{n}})\] Also heavier the nucleus, more is the mass defect Thus, \[20({{m}_{n}}+{{m}_{p}})-{{M}_{2}}>10({{m}_{p}}+{{m}_{n}})-{{M}_{1}}\] or \[10({{m}_{p}}+{{m}_{n}})>{{M}_{2}}-{{M}_{1}}\] or \[{{M}_{2}}<10({{m}_{p}}+{{m}_{n}})+{{M}_{1}}\] Now, since \[{{M}_{1}}<10({{m}_{p}}+{{m}_{n}})\]
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