Correct answer is 5

To determine the rate of converting ${}^4\mathrm{He}$ to ${}^{12}\mathrm{C}$, we need to calculate the energy released per reaction and use the given power production of the star to find the rate of reactions. The relevant nuclear reaction is:

${}^4\mathrm{He}+{}^4\mathrm{He}+{}^4\mathrm{He}\to {}^{12}\mathrm{C}+\mathrm{Q}$

The masses involved in the reaction are given:

• Mass of ${}^4\mathrm{He}=4.0026\mathrm{u}$
• Mass of ${}^{12}\mathrm{C}=12\mathrm{u}$

First, we calculate the mass defect (difference between the mass of reactants and products) which will give us the energy released in each reaction:

Mass of reactants: $3\times 4.0026\mathrm{u}=12.0078\mathrm{u}$

Mass of product: $12\mathrm{u}$

Mass defect: $12.0078\mathrm{u}-12\mathrm{u}=0.0078\mathrm{u}$

We use Einstein's mass-energy equivalence principle, $E=m{c}^2$, to find the energy released per reaction. The conversion factor between atomic mass units and energy is $1\mathrm{u}=931.5\mathrm{MeV}$.

Energy released per reaction: $0.0078\mathrm{u}\times 931.5\mathrm{MeV}/\mathrm{u}=7.2627\mathrm{MeV}$

We convert this energy into joules. $1\mathrm{MeV}=1.60218\times {10}^{-13}\mathrm{J}$:

Energy per reaction: $7.2627\mathrm{MeV}\times 1.60218\times {10}^{-13}\mathrm{J}/\mathrm{MeV}=1.163\times {10}^{-12}\mathrm{J}$

The power generated by the star is given as $5.808\times {10}^{30}\mathrm{W}$. The rate of the reaction is the power divided by the energy per reaction:

$\text{Rate of reactions}=\frac{\text{Power}}{\text{Energy per reaction}}$

$\text{Rate}=\frac{5.808\times {10}^{30}\mathrm{W}}{1.163\times {10}^{-12}\mathrm{J}}$

$\text{Rate}=4.99\times {10}^{42}{\mathrm{s}}^{-1}\simeq 5\times {10}^{42}{\mathrm{s}}^{-1}$

Thus, the rate of converting ${}^4\mathrm{He}$ to ${}^{12}\mathrm{C}$ is:

$\mathrm{n}=5$

Correct answer is 727

To find the energy released when three helium nuclei combine to form a carbon nucleus, we first need to understand that this process is essentially nuclear fusion, forming a heavier nucleus from lighter ones. The mass defect in this fusion process is the key to calculating the energy released, according to Einstein's equation $E=\mathrm{\Delta }m{c}^2$, where $E$ is the energy released, $\mathrm{\Delta }m$ is the mass defect, and $c$ is the speed of light.

The atomic mass of a helium nucleus (also known as an alpha particle) is $4.002603\text{u}$.

1. Calculate the total initial mass of three helium nuclei:

$\text{Total initial mass}=3\times 4.002603\text{u}=12.007809\text{u}$

2. The atomic mass of a carbon nucleus formed by the fusion of three helium nuclei is not directly given, but we can infer it's approximately $12\text{u}$, based on knowledge of isotopes and considering that the question appears to simplify the carbon nucleus to a mass number of 12 (common carbon-12 isotope).

3. Calculate the mass defect ($\mathrm{\Delta }m$):

$\mathrm{\Delta }m=\text{Total initial mass}-\text{Final mass}$

$\mathrm{\Delta }m=12.007809\text{u}-12\text{u}=0.007809\text{u}$

4. Convert the mass defect to energy. Given $1\text{u}=931{\text{MeV/c}}^2$, the energy released is calculated using the formula $E=\mathrm{\Delta }m{c}^2$:

$E=0.007809\text{u}\times 931\text{MeV/u}=7.271839\text{MeV}$

Since the question asks for the answer in the format of $\times {10}^{-2}\text{MeV}$, we convert the energy released:

$7.271839\text{MeV}=727.1839\times {10}^{-2}\text{MeV}$

Therefore, the energy released in this reaction is approximately $727.1839\times {10}^{-2}\text{MeV}$. The exact value might differ slightly depending on how the atomic mass of the carbon nucleus is considered or rounded in specific scenarios, but based on the information provided, this is a suitable approximation.

Correct answer is 208

Q. value

$\begin{array}{rl}& =\{(235.0439)-[39.9054+93.9063+1.0086]\}\times 931\mathrm{MeV}\\ & =208\mathrm{MeV}\end{array}$

Correct answer is 1

$\begin{array}{rl}& \mathrm{E}=\mathrm{\Delta }{\mathrm{mc}}^2\\ & =0.4\times {10}^{-3}\times {(3\times {10}^8)}^2\\ & =3600\times {10}^7\mathrm{kWs}\\ & =\frac{3600\times {10}^7}{3600}\mathrm{kWh}=1\times {10}^7\mathrm{kWh}\end{array}$

Correct answer is 236

To determine the energy released in a nuclear fission process, we use the difference in binding energy (BE) before and after the fission. The formula for energy released ($Q$ value) in the process is given by:

$Q=(\text{Total BE of products})-(\text{Total BE of reactants})$

In this case, the reactant is a high mass nuclide with atomic mass $A\approx 236$ and a binding energy of $7.6\mathrm{MeV}/\mathrm{nucleon}$. Each of the two middle mass nuclides formed as products has atomic mass $A\approx 118$ and a binding energy of $8.6\mathrm{MeV}/\mathrm{nucleon}$.

Therefore, we calculate the total binding energy of reactant and products as follows:

For reactant:

${\text{BE}}_{\text{reactant}}=236\times 7.6\mathrm{MeV}$

For products (since there are two identical products):

${\text{BE}}_{\text{products}}=2\times (118\times 8.6)\mathrm{MeV}$

Thus, the energy released ($Q$ value) is:

$Q={\text{BE}}_{\text{products}}-{\text{BE}}_{\text{reactant}}$

$Q=2(118\times 8.6)-(236\times 7.6)$

$Q=236\times (8.6-7.6)$

$Q=236\times 1$

$Q=236\mathrm{MeV}$

This calculation demonstrates how the difference in binding energy per nucleon before and after fission leads to the release of energy, consistent with the mass-energy equivalence principle.