Question

The ratio of area of first excited state to ground state of orbit of hydrogen atom is

• A. 1:16
• B. 1:4
• C. 4:1
• D. 16:1

Answer 1

The Correct Option is D

In a hydrogen atom, the area of the orbit is proportional to the square of the principal quantum number \( n \). The formula for the radius of an orbit is given by: \[ r_n = n^2 \cdot r_1 \] where \( r_1 \) is the radius of the ground state (\( n = 1 \)). The area of the orbit \( A_n \) is proportional to \( r_n^2 \), so the ratio of the areas of the first excited state (\( n = 2 \)) and the ground state (\( n = 1 \)) is: \[ \frac{A_2}{A_1} = \frac{(2^2 \cdot r_1)^2}{(1^2 \cdot r_1)^2} = \frac{4^2}{1^2} = 16 \] Thus, the ratio of the area of the first excited state to the ground state is \( 16:1 \). Thus, the solution is \( 16:1 \). 

Answer 2

In the Bohr model of the hydrogen atom, the radius of the electron's orbit in the \(n\)-th energy level is given by: \[ r_n = n^2 r_1 \] where \(r_1\) is the radius of the ground state (\(n=1\)), and \(n\) is the principal quantum number. - For the ground state, \(n = 1\), so the radius is \(r_1\). - For the first excited state, \(n = 2\), so the radius is \(r_2 = 2^2 r_1 = 4r_1\). Now, the area \(A\) of the orbit is proportional to the square of the radius of the orbit. Hence, the areas of the orbits are given by: \[ A_n \propto r_n^2 \] - The area of the ground state orbit (\(A_1\)) is proportional to \(r_1^2\). - The area of the first excited state orbit (\(A_2\)) is proportional to \(r_2^2 = (4r_1)^2 = 16r_1^2\). Thus, the ratio of the areas of the first excited state to the ground state is: \[ \frac{A_2}{A_1} = \frac{16r_1^2}{r_1^2} = 16 \] Therefore, the ratio of the area of the first excited state to the ground state is 16:1. The correct answer is (D).