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Nuclei — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsNuclei4 Marks
Question
When subatomic particles undergo reactions, energy is conserved, but mass is not necessarily conserved. However, a particle's mass “contributes” to its total energy, in accordance with Einstein's famous equation, $E = mc^2$ In this equation, E denotes the energy carried by a particle because of its mass. The particle can also have additional energy due to its motion and its interactions with other particles. Consider a neutron at rest and well separated from other particles. It decays into a proton, an electron and an undetected third particle as given here: Neutron → proton + electron + ???
The given table summarizes some data from a single neutron decay. Electron volt is a unit of energy. Column 2 shows the rest mass of the particle times the speed of light squared.
Particle
$Mass\times c^2$ (MeV)
Kinetic energy (MeV)
Neutron
940.97
0.00
Proton
939.67
0.01
Electron
0.51
0.39
  1. From the given table, which properties of the undetected third particle can be calculate?
  1. Total energy, but not kinetic energy.
  2. Kinetic energy, but not total energy.
  3. Both total energy and kinetic energy.
  4. Neither total energy nor kinetic energy.
  1. Assuming the table contains no major errors, what can we conclude about the $(mass \times c^2)$ of the undetected third particle?
  1. It is 0. 79 MeV
  2. It is 0.39 MeV
  3. It is less than or equal to 0.79 MeV; but we cannot be more precise.
  4. It is less than or equal to 0.40 MeV; but we cannot be more precise.
  1. Could this reaction occur?
Proton → neutron + other particles
  1. Yes, if the other particles have much more kinetic energy than mass energy.
  2. Yes, but only if the proton has potential energy (due to interactions with other particles).
  3. No, because a neutron is more massive than a proton.
  4. No, because a proton is positively charged while a neutron is electrically neutral.
  1. How much mass has to be converted into energy to produce electric power of 500MW for one hour?
  1. $2 \times 10^{-5}kg$
  2. $1 \times 10^{-5}kg$
  3. $3 \times 10^{-5}kg$
  4. $4 \times 10^{-5}kg$
  1. The equivalent energy of 1g of substance is.
  1. $9 \times 10^{13}J$
  2. $6 \times 10^{12}J$
  3. $3 \times 10^{13}J$
  4. $6 \times 10^{13}J$
✓

Answer

  1. (a) Total energy, but not kinetic energy.
Explanation:
As just shown, energy conservation allows us to calculate the third particle's total energy. But we do not know what percentage of that total is mass energy.
  1. (d) It is less than or equal to 0.40 MeV, but we cannot be more precise.
Explanation:
According to the passage, subatomic reactions do not conserve mass. So, we cannot find the third particle's mass by setting $\text{m}_\text{neutron}$equal to $\text{m}_\text{proton}+\text{m}_\text{electron}+\text{E}_\text{thrid particle}$
The neutron has energy 940.97 MeV. The proton has energy 939.67 MeV + 0.01 MeV = 939.69 MeV. The electron has energy 0.51 MeV + 0.39 MeV = 0.90 MeV. Therefore, the third particle has energy.
$\text{E}_\text{third particle}=\text{E}_\text{neutron}-\text{E}_\text{proton}-\text{E}_\text{electron}$
$= 940.97 - 939.67 - 0.90 = 0.40$ MeV
We just found the third particle's total energy, the sum of its mass energy and kinetic energy. Without more information, we cannot figure out how much of that energy is mass energy.
  1. (b) Yes, but only if the proton has potential energy (due to interactions with other particles).
  2. (a) $2 \times 10^{-5}kg$
Explanation:
Here, $P = 500MW = 5 \times 10^8W,$
$t = 1h = 3600s$
Energy produced, $E = P \times t = 5 \times 10^8 \times 3600 = 18 \times 10^{11}J$
$\text{As}\ \text{E}=\Delta\text{mc}^2$
$\therefore\Delta\text{m}=\frac{\text{E}}{\text{c}^2}=\frac{18\times10^{11}}{(3\times10^8)^2}$
$=\frac{18\times10^{11}}{(3\times10^8)^{16}}=2\times10^{-5}\text{kg}$
  1. (a) $9 \times 10^{13}J$
Explanation:
Using, $E = mc^2$
Here, $m = 1g = 1 \times 10^{-3}kg, c = 3 \times 10^8m s^{-1}$
$\therefore E = 10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13}J$

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