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Question: What is excess mass \({\Delta _1}\) of \(^1H\) (actual mass is \(1.007825u\) ) in (a) atomic mass un...

What is excess mass Δ1{\Delta _1} of 1H^1H (actual mass is 1.007825u1.007825u ) in (a) atomic mass units and (b) MeVc−2MeV{c^{ - 2}} ? What is the mass excess of Δn{\Delta _n} of a neutron (actual mass is 1.008665u1.008665u ) in (c) atomic mass units and (d) MeVc−2MeV{c^{ - 2}} ? What is mass excess Δ120{\Delta _{120}} of 120Sn^{120}Sn (actual mass is 119.902197u119.902197u ) in (e) atomic mass units and (f) MeVc−2MeV{c^{ - 2}} ?

Explanation

Solution

In order to solve this question we need to understand Nuclear reaction. So a nuclear reaction is defined as the reaction in which either heavier nuclei disintegrate or lighter nuclei combine to form heavier nucleus, however in both the cases there is some mass defect and due to this there is heat energy released during reaction.

Complete step by step answer:
Nuclear reaction is of two types one is nuclear fission and fusion. Fission is defined as a nuclear reaction in which two small nuclei combine to form a large nucleus and thereby radiating heat. Nuclear fusion is defined as a nuclear reaction in which a large nucleus breaks into a small nucleus and also some particles like neutrino and antineutrino etc. Mass excess is defined as difference between actual mass and atomic mass given by,
Δ=M0−MA\Delta = {M_0} - {M_A}
And energy from mass excess is given by,
E=Δ×931.5MeVc−2E = \Delta \times 931.5MeV{c^{ - 2}}

So for (a) We have, MA=1u{M_A} = 1u and M0=1.007825u{M_0} = 1.007825u
So Δ=(1.007825−1)u\Delta = (1.007825 - 1)u
Δ=0.007825u\Delta = 0.007825u
For (b) Energy is given by, E=0.007825×931.5MeVc−2E = 0.007825 \times 931.5MeV{c^{ - 2}}
E=+7.290MeVc−2E = + 7.290MeV{c^{ - 2}}
Similarly for (c) we have, MA=1u{M_A} = 1u and M0=1.008665u{M_0} = 1.008665u
So Δ=(1.008665−1)u\Delta = (1.008665 - 1)u
Δ=0.008665u\Delta = 0.008665u

For (d) Energy is given by,
E=0.008665×931.5MeVc−2E = 0.008665 \times 931.5MeV{c^{ - 2}}
⇒E=+8.071MeVc−2\Rightarrow E = + 8.071MeV{c^{ - 2}}
Similarly for (e) we have,
MA=120u{M_A} = 120u and M0=119.9021u{M_0} = 119.9021u
So Δ=(119.9021−120)u\Delta = (119.9021 - 120)u
Δ=−0.09780u\Delta = - 0.09780u
For (f) Energy is given by,
E=−0.0978×931.5MeVc−2E = - 0.0978 \times 931.5MeV{c^{ - 2}}
∴E=−91.1 MeVc−2\therefore E = - 91.1\,MeV{c^{ - 2}}

Note: It should be remembered that nuclear reactions only occur if the nuclei are radioactive in nature. Radioactivity decay is the process in which an unstable atomic nucleus loses energy by radiation. There are three types of rays which emanate from the nucleus, one is alpha ray which is nucleus of helium it is less penetrating in power, next is beta radiation in which electron and positron emanate from nucleus and gamma radiation which is highly penetrating in nature but it has no charge.