COMEDK 2025 Morning Shift | Atoms and Nuclei Question 13 | Physics

In a nuclear fusion reaction, two nuclei, $A$ and $B$ fuse to produce a nucleus $C$, releasing an amount of energy $\Delta \mathrm{E}$ in the process. If the mass defects of the three nuclei are $\Delta M_A, \Delta M_B$ and $\Delta M_C$ respectively, then which of the following relations is true? ( $c$ is the speed of light).

$\Delta M_A+\Delta M_B=\Delta M_C+\frac{\Delta E}{c^2}$

$\Delta M_A-\Delta M_B=\Delta M_C+\frac{\Delta E}{c^2}$

$\Delta M_A-\Delta M_B=\Delta M_C-\frac{\Delta E}{c^2}$

$\Delta M_A+\Delta M_B=\Delta M_C-\frac{\Delta E}{c^2}$

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

The binding energy per nucleon for $$\mathrm{C}^{12}$$ is $$7.68 \mathrm{~MeV}$$ and that for $$\mathrm{C}^{13}$$ is $$7.47 \mathrm{~MeV}$$. The energy required to remove a neutron from $$\mathrm{C}^{13}$$ is

$$7.92 \times 10^{-13} \mathrm{~MeV}$$

$$4.95 \times 10^{-13} \mathrm{eV}$$

$$7.92 \times 10^{-13} \mathrm{~J}$$

$$7.92 \times 10^{-19} \mathrm{~J}$$

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

The distance of closest approach when an alpha particle of kinetic energy $$6.5 \mathrm{~MeV}$$ strikes a nucleus of atomic number 50 is

0.221 fm

1.101 fm

0.0221 fm

4.42 fm

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength $$\lambda$$. When it jumps from the second to the first orbit, the corresponding wavelength of the photon will be

$$ \frac{5 \lambda}{27} $$

$$ \frac{7 \lambda}{20} $$

$$ \frac{16 \lambda}{9} $$

$$ \frac{20 \lambda}{7} $$

PREVIOUS NEXT

COMEDK Subjects