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

Answer: (D)

n = 2

Answer 2

r = (0.529 Å) x n2 / Z
Where r is the radius, n is the principal quantum number, and Z is the atomic number.
Given that the radius changes from 0.53 Å to 2.1 Å, we can set up the following equation:
2.1 = (0.529) x n2 / Z
To determine the change in the principal quantum number, we can rearrange the equation:
n2 = (2.1 x Z) / 0.529
Since Z for a hydrogen atom is 1, we can simplify the equation to:
n2 = 2.1 / 0.529
n2 = 3.974
Taking the square root of both sides, we find:
n ≈ 1.993
Since the principal quantum number must be a whole number, the value of n that satisfies the equation is n = 2.
Therefore, the correct option is (D) n = 2.