NEET UG PHYSICS MCQ: NUCLEI

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Application +4 | -1

In the reaction

_1^2H + \,_1^3\,\,H \to _2^4He + \,_0^1\,\,n

if the binding energies of

_1^2H,_1^3\,H

and

_2^4He

are respectively

a,b

and

c

(in

MeV

), then the energy (in

MeV

) released in this reaction is

View answer and explanation

Correct option: B

c - a - b

Option B (

c - a - b

) is correct because the energy released is the difference between the binding energy of the products and the reactants.

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\mathrm{216}

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{ }_{92} X^{236} \rightarrow{ }_{56} Y^{141}+{ }_{36} Z^{92}+3 R

({R})

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\left({ }_0^1 \mathrm{n}\right)

\left({ }_{92}^{235} \mathrm{U}\right)

1\,\,a.m.u = 931\,\,MeV