Correct answer option is (A)

In nuclear disintegration, the conservation of momentum plays a crucial role in determining the motion of the resulting fragments. Since the original nucleus is at rest, its total initial momentum is zero. After the disintegration, the total momentum of the system must still be zero to conserve momentum.

Given the mass ratio of the resulting two smaller nuclei is $2:1$, let's denote the masses of the two nuclei as $2m$ and $m$, respectively.

Applying Conservation of Momentum

For the nucleus with mass $2m$ and velocity $v_1$ and for the nucleus with mass $m$ and velocity $v_2$, the conservation of momentum equation is:

$2m\cdot v_1+m\cdot v_2=0$

Given that momentum is a vector quantity, and the total initial momentum was zero, the nuclei must move in opposite directions for their momenta to cancel each other out. Therefore, we rearrange the equation:

$2m\cdot v_1=-m\cdot v_2$

$2\cdot v_1=-v_2$

$v_2=-2\cdot v_1$

The negative sign indicates that $v_1$ and $v_2$ are in opposite directions. Taking magnitudes and considering the ratio:

$\left|v_2\right|=2\cdot \left|v_1\right|$

This equation shows that the velocity of the lighter fragment (mass $m$) is twice the velocity of the heavier fragment (mass $2m$). Since they must move in opposite directions for momentum conservation, this confirms:

Correct Answer:

Option A - in opposite directions with speed in the ratio of $1:2$ respectively.

Correct answer option is (A)

To determine the ratio $\frac{E_H}{E_U}$, let's first consider the energy released in the fusion of hydrogen and the energy released in the fission of ${}^{235}U$.

For the fusion reaction:

The given reaction is:

$4_1^{1}H+2{\mathrm{e}}^{-}\to {}_2^4\mathrm{He}+2\nu +6\gamma +26.7\mathrm{MeV}$

This reaction shows that 4 hydrogen atoms and 2 electrons produce 1 helium atom, 2 neutrinos, and 6 gamma photons while releasing 26.7 MeV of energy.

The mass of 1 mole of $H_1^1$ is approximately 1 gram, so 2 kg of hydrogen is equal to:

$2\times {10}^3\mathrm{g}$

The number of moles of hydrogen in 2 kg is:

${\mathrm{N}}_{\text{moles}}=\frac{2\times {10}^3}{1}=2\times {10}^3\mathrm{moles}$

The number of hydrogen atoms in 2 kg is:

${\mathrm{N}}_{\text{atoms}}=N_A\times 2\times {10}^3=6.023\times {10}^{23}\times 2\times {10}^3=1.2046\times {10}^{27}\mathrm{atoms}$

Since 4 hydrogen atoms release 26.7 MeV, the total energy released ($E_H$) is:

$E_H=\frac{1.2046\times {10}^{27}}{4}\times 26.7\mathrm{MeV}$

$E_H=3.0115\times {10}^{26}\times 26.7\mathrm{MeV}$

$E_H\approx 8.04\times {10}^{27}\mathrm{MeV}$

For the fission reaction:

The energy released per fission of one ${}^{235}U$ nucleus is 200 MeV.

The mass of 1 mole of ${}^{235}U$ is approximately 235 grams, so 2 kg of ${}^{235}U$ is equal to:

$2\times {10}^3\mathrm{g}$

The number of moles of ${}^{235}U$ in 2 kg is:

${\mathrm{N}}_{\text{moles}}=\frac{2\times {10}^3}{235}\approx 8.51\mathrm{moles}$

The number of ${}^{235}U$ nuclei is:

${\mathrm{N}}_{\text{nuclei}}=N_A\times 8.51\approx 6.023\times {10}^{23}\times 8.51\approx 5.12\times {10}^{24}\mathrm{nuclei}$

The total energy released ($E_U$) is:

$E_U=200\mathrm{MeV}\times 5.12\times {10}^{24}\approx 1.024\times {10}^{27}\mathrm{MeV}$

Finally, the ratio $\frac{E_H}{E_U}$ is:

$\frac{E_H}{E_U}\approx \frac{8.04\times {10}^{27}}{1.024\times {10}^{27}}\approx 7.85$

Therefore, the ratio is closest to option A: 7.62.

Correct answer option is (A)

To determine the energy equivalent of a mass, we use Einstein's mass-energy equivalence principle given by the equation:

$E=m{c}^2$

where:

- $E$ is the energy

- $m$ is the mass

- $c$ is the speed of light in a vacuum, which is approximately $3\times {10}^8\mathrm{m}/\mathrm{s}$

Given the mass $m=1\mathrm{g}=1\times {10}^{-3}\mathrm{kg}$, we can substitute these values into the equation:

$E=(1\times {10}^{-3}\mathrm{kg})\times (3\times {10}^8\mathrm{m}/\mathrm{s}{)}^2$

Calculating this, we get:

$E=1\times {10}^{-3}\times 9\times {10}^{16}$

$E=9\times {10}^{13}\mathrm{J}$

Next, to convert this energy into electron volts ($\mathrm{eV}$), we use the conversion factor: $1\mathrm{J}=6.242\times {10}^{12}\mathrm{MeV}$.

Therefore:

$E=9\times {10}^{13}\mathrm{J}\times 6.242\times {10}^{12}\mathrm{MeV}/\mathrm{J}$

Calculating this, we get:

$E=5.6178\times {10}^{26}\mathrm{MeV}$

Therefore, the energy equivalent of $1\mathrm{g}$ of a substance is:

Option A: $5.6\times {10}^{26}\mathrm{MeV}$

Correct answer option is (A)

To determine the nuclear binding energy of the isotope ${}_5^{12}B$, we need to consider the mass defect concept. The mass defect is the difference between the sum of the individual masses of nucleons (protons and neutrons) and the actual mass of the nucleus.

Let's calculate the mass defect first. The isotope ${}_5^{12}B$ has 5 protons and 7 neutrons (since the total number of nucleons is 12). Therefore, the mass defect can be written as:

$\mathrm{\Delta }M=(5M_p+7M_n)-M_0$

Once we have the mass defect, the binding energy can be found using Einstein's mass-energy equivalence principle, which is given by:

$E=\mathrm{\Delta }M\cdot c^2$

Substituting the mass defect into this equation, we get:

$E=((5M_p+7M_n)-M_0)\cdot c^2$

Therefore, the correct answer is Option A:

$(5M_p+7M_n-M_0)\cdot c^2$

Correct answer option is (B)

In the given hypothetical fission reaction:

${}_{92}^{236}X\to {}_{56}^{141}Y+{}_{36}^{92}Z+3R$

We need to determine the identity of particles denoted by $R$. Let's use the conservation of charge and mass number (nucleon number) to identify $R$.

First, for the conservation of nucleon number (mass number), we have:

$236=141+92+3\times A_R$

Where $A_R$ is the mass number of $R$. This simplifies to:

$236=233+3A_R$

$3A_R=236-233$

$3A_R=3$

$A_R=1$

Next, we use the conservation of charge (atomic number), we have:

$92=56+36+3Z_R$

Where $Z_R$ is the atomic number of $R$. This simplifies to:

$92=92+3Z_R$

$3Z_R=92-92$

$3Z_R=0$

$Z_R=0$

Since the particle $R$ has a mass number of 1 and atomic number of 0, it must be a neutron.

So, the identity of the emitted particles $R$ is:

Option B: Neutron