Find the gravitational attraction between the two atoms in a hydrogen molecule? distance between the two atoms 1x 10−10 m.

Find the gravitational attraction between the two atoms in a hydrogen molecule. Given: \( G = 6.67 \times 10^{-11} \, \text{Nm}^2 \, \text{kg}^{-2} \), mass of a hydrogen atom = \( 1.67 \times 10^{-27} \, \text{kg} \), and distance between the two atoms = \( 1 \times 10^{-10} \, \text{m} \).

A) \( 1.85 \times 10^{-47} \, \text{N} \)
B) \( 2.23 \times 10^{-46} \, \text{N} \)
C) \( 1.87 \times 10^{-44} \, \text{N} \)
D) \( 3.21 \times 10^{-45} \, \text{N} \)

Answer: A) \( 1.85 \times 10^{-47} \, \text{N} \)

Solution:

The gravitational force between two masses is given by:

\[ F = G \frac{m_1 m_2}{r^2} \]

Here:

• \( G = 6.67 \times 10^{-11} \, \text{Nm}^2 \, \text{kg}^{-2} \)
• \( m_1 = m_2 = 1.67 \times 10^{-27} \, \text{kg} \)
• \( r = 1 \times 10^{-10} \, \text{m} \)

Substituting these values:

\[ F = 6.67 \times 10^{-11} \cdot \frac{(1.67 \times 10^{-27}) \cdot (1.67 \times 10^{-27})}{(1 \times 10^{-10})^2} \]

Simplifying:

\[ F = 6.67 \times 10^{-11} \cdot \frac{2.79 \times 10^{-54}}{1 \times 10^{-20}} \]

\[ F = 6.67 \times 10^{-11} \cdot 2.79 \times 10^{-34} \]

\[ F = 1.85 \times 10^{-47} \, \text{N} \]

Final Answer: The gravitational attraction between the two hydrogen atoms is \( 1.85 \times 10^{-47} \, \text{N} \).

MCQ On Motion In Plane | Newton's Laws Of Motion | CBSE, IIT JEE, NEET, CET | Physics PYQ

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Find the gravitational attraction between the two atoms in a hydrogen molecule. Given: \( G = 6.67 \times 10^{-11} \, \text{Nm}^2 \, \text{kg}^{-2} \), mass of a hydrogen atom = \( 1.67 \times 10^{-27} \, \text{kg} \), and distance between the two atoms = \( 1 \times 10^{-10} \, \text{m} \).