QJEEM24Phf3bf: JEE Main 2024 Physics - Class 12 JEEM24Phf3bf Solution: In hydrogen like system the ratio of coulombian force and…

To find the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system, we use the formulae for both forces and then divide them.

The Coulombic (electrostatic) force, $F_C$ , between two charges is given by Coulomb's law:

$F_C = k \frac{|q_1 q_2|}{r^2}$

where $k$ is Coulomb's constant ( $8.987 \times 10^9 \, \text{Nm}^2\text{C}^{-2}$ ), $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between the charges. For a proton and an electron, $q_1 = q_2 = e$ , where $e$ is the elementary charge ( $1.602 \times 10^{-19} \, \text{C}$ ).

The gravitational force, $F_G$ , between two masses is given by Newton's law of universal gravitation:

$F_G = G \frac{m_1 m_2}{r^2}$

where $G$ is the gravitational constant ( $6.674 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2}$ ), and $m_1$ and $m_2$ are the masses of the two objects. For a proton and an electron, $m_p \approx 1.673 \times 10^{-27} \, \text{kg}$ and $m_e \approx 9.109 \times 10^{-31} \, \text{kg}$ , respectively.

The ratio of the Coulombic to the gravitational force is therefore:

$\frac{F_C}{F_G} = \frac{k \frac{|e^2|}{r^2}}{G \frac{m_p m_e}{r^2}} = \frac{k e^2}{G m_p m_e}$

Plugging in the values:

$\frac{F_C}{F_G} = \frac{(8.987 \times 10^9) (1.602 \times 10^{-19})^2}{(6.674 \times 10^{-11}) (1.673 \times 10^{-27}) (9.109 \times 10^{-31})}$

$\frac{F_C}{F_G} = \frac{(8.987 \times 10^9) \times (2.568 \times 10^{-38})}{(6.674 \times 10^{-11}) \times (1.523 \times 10^{-57})}$

$\frac{F_C}{F_G} \approx 2.3 \times 10^{39}$

Therefore, the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system is on the order of $10^{39}$ , making Option B the correct answer.