Question

The radius of hydrogen atom in the ground state is 0.53 Å .After collision with an electron it is found to have a radius of 2.1 Å, the principal quantum number 'n' of the state of the atom is 

• A. n = 3
• B. n=1
• C. n = 4
• D. n = 2

Answer 1

The Correct Option is D

The radius of the orbit is given by the formula: 

\( r = \frac{0.529 \, \text{Å} \times n^2}{Z} \)

Where:

• r = radius of the orbit
• n = principal quantum number
• Z = atomic number

Given that the radius changes from 0.53 Å to 2.1 Å, we set up the following equation:

\( 2.1 = \frac{0.529 \times n^2}{Z} \)

To determine the change in the principal quantum number, we rearrange the equation:

\( n^2 = \frac{2.1 \times Z}{0.529} \)

For a hydrogen atom, \( Z = 1 \), so the equation simplifies to:

\( n^2 = \frac{2.1}{0.529} \)

Calculating the right-hand side:

\( n^2 = 3.974 \)

Taking the square root of both sides, we find:

\( n \approx 1.993 \)

Since the principal quantum number must be a whole number, the value of \( n \) that satisfies the equation is:

\( n = 2 \)

Answer 2

The relationship between the radius \( r \) and the principal quantum number \( n \) is given by:

\( r \propto n^2 \) 

Therefore, the ratio of the radii \( r_2 \) and \( r_1 \) is:

\[ \frac{r_2}{r_1} = \left( \frac{n_2}{n_1} \right)^2 \]

Substitute the given value \( \frac{r_2}{r_1} = 0.25 \):

\[ 0.25 = \left( \frac{n_2}{n_1} \right)^2 \]

Solving for \( n_2 \), we get:

\[ n_2^2 = 0.25 \times n_1^2 \]

Since \( n_1 = 1 \), the equation becomes:

\[ n_2^2 = 0.25 \times 1^2 = 0.25 \]

Thus, taking the square root of both sides:

\[ n_2 = 2 \]

Therefore, the value of \( n_2 \) is \( 2 \).