Find the energy released in the reaction 1H2+ 1H2 overset arrow 2He4 if binding energy per nucleon for deuteron and α particle are x1 and x2, respectively.

Question

Find the energy released in the reaction 1H2+ 1H2 overset arrow 2He4 if binding energy per nucleon for deuteron and α particle are x1 and x2, respectively. (A) 4(.x2-x1.) (B) 4(.x1+x2.) (C) (.x1 -x2.) (D) (.x1 +x2.)

Tardigrade Admin · Accepted Answer

The binding energy on reactant side ( i.e., deutrons) is 4× x1=4x1 The binding energy on product side ( i.e., α -particle) is 4× x2=4x2 So, energy released Q is 4x2-4x1=4(x2 - x1)

Q. Find the energy released in the reaction $_1 H_{2} +_1 H_{2} \to_2 H e^{4}$ if binding energy per nucleon for deuteron and $α$ particle are x₁and x₂, respectively.

$4 (x_{2} - x_{1})$

$4 (x_{1} + x_{2})$

$(x_{1} - x_{2})$

$(x_{1} + x_{2})$

The binding energy on reactant side ( i.e., deutrons) is $4 \times x_{1} = 4 x_{1}$
The binding energy on product side ( i.e., $α$ -particle) is $4 \times x_{2} = 4 x_{2}$
So, energy released $Q$ is $4 x_{2} - 4 x_{1} = 4 (x_{2} - x_{1})$

Q. Find the energy released in the reaction _1H2​+_1H2​→_2He4 if binding energy per nucleon for deuteron and α particle are x1 and x2, respectively.

4(x2​−x1​)

4(x1​+x2​)

(x1​−x2​)

(x1​+x2​)