Question

Find the Q value of the reaction. The mass of 7/3Li is 7.016 u, the mass of 4/2He is 4.0026 u, the mass of the product is 10.0129 u and the mass of the neutron is 1.00866 u .

1. −2.75724 MeV

2. −123.099 MeV 3.

3.09877 MeV

4. −28.0977 MeV

5. 5.09873 MeV

6. 32.0987 MeV

Answer 1

Given that,

⁷Li₃ = 7.016 u

⁴He₂ = 4.0026 u

Product mass = 10.0129 u and neutron mass = 1.00866 u

7.016 + 4.0026 = 11.0186 u

There is a mass defect =11.0186 - (10.0129 + 1.00866) = - 0.00296 u

We know that, 1 amu = 1.661 x 10^-27 kg, So - 0.00296 amu = - 4.915 x 10^-30 kg

We know that,

E = m c² = - 4.915 x 10^-30 kg x (3 x 10⁸ m/s)² = - 44.235 x 10^-14 Joules

Now 1 MeV = 1.602 x 10^-13 Joules

So, E = - 2.76 MeV This is approx equal to option 1.

Hence option 1 E = -2.75724 MeV is the correct answer.