Question

find the energy released when $_{82}^{211}pb$ undergoes $\beta$ decay to become $_{83}^{211}bi$. be sure to take into account the mass of the electrons associated with the neutral atoms.

a. 7.2mev
b. 10.4mev
c. 25.8mev
d. 1.3mev
e. 5.1mev

Explanation:

Step1: Write the beta - decay equation

The beta - decay of $^{211}_{82}Pb$ to $^{211}_{83}Bi$ is $^{211}_{82}Pb
ightarrow^{211}_{83}Bi + e^-+\bar{
u}_e$. The energy released $Q$ in a nuclear decay is given by the mass - defect $\Delta m$ times $c^{2}$, $Q = \Delta mc^{2}$. The mass of $^{211}_{82}Pb$ is $m_{Pb}=210.988731\ u$ and the mass of $^{211}_{83}Bi$ is $m_{Bi}=210.987255\ u$. The mass of an electron $m_e = 0.00054858\ u$.

Step2: Calculate the mass - defect

The mass - defect $\Delta m=m_{Pb}-m_{Bi}-m_e$. Substituting the values: $\Delta m = 210.988731\ u-210.987255\ u - 0.00054858\ u=0.00092742\ u$.

Step3: Convert mass - defect to energy

We know that $1\ u = 931.5\ MeV/c^{2}$. So, $Q=\Delta m\times931.5\ MeV/u$. Substituting $\Delta m = 0.00092742\ u$ into the formula, we get $Q=0.00092742\times931.5\ MeV\approx 0.864\ MeV$. However, if we consider the electron - capture equivalent way and account for the electron masses in a more comprehensive way (since in the neutral - atom mass, the electron masses are already included in the atomic masses), the correct mass - defect calculation for beta - decay gives $\Delta m=(m_{Pb}-m_{Bi})$. $\Delta m=(210.988731 - 210.987255)\ u= 0.001476\ u$. Then $Q=\Delta m\times931.5\ MeV/u=0.001476\times931.5\ MeV\approx1.37\ MeV\approx1.3\ MeV$.

Answer:

d. $1.3\ MeV$