Question

A 74^Be atom (mass = 7.0169 amu) decays into a 73^Li atom (mass = 7.0160 amu) by electron capture. How much energy (in millions of electron volts, MeV) is produced by this reaction?

a) 0.0009 MeV
b) 0.092 MeV
c) 0.19 MeV
d) 0.90 MeV

Answer 1

The energy released by the decay of a 74Be atom into a 73Li atom by electron capture can be calculated using the equation E = (m_initial - m_final)c^2. On calculation, the energy released is found to be 0.8396 MeV.

Step-by-step explanation:

The energy released by the decay of a 74Be atom into a 73Li atom by electron capture can be calculated using the equation E = (minitial - mfinal)c2 where E is the energy released, minitial is the initial mass, mfinal is the final mass, and c is the speed of light.

Given that the initial mass is 7.0169 amu and the final mass is 7.0160 amu, we can calculate the energy released as follows:

1. Calculate the change in mass: Δm = minitial - mfinal = 7.0169 amu - 7.0160 amu = 0.0009 amu
2. Convert the change in mass to kilograms: Δm = 0.0009 amu x (1.66 x 10-27 kg/amu) = 1.494 x 10-30 kg
3. Calculate the energy released: E = (1.494 x 10-30 kg) x (3 x 108 m/s)2 = 1.3446 x 10-13 J
4. Convert the energy from joules to MeV: 1 J = 6.242 x 1018 MeV, so the energy released is 1.3446 x 10-13 J x (6.242 x 1018 MeV/J) = 0.8396 MeV

Therefore, the amount of energy produced by this reaction is 0.8396 MeV, which is closest to option d) 0.90 MeV.