Question

The Bohr orbit radius for the hydrogen atom (\(n = 1\)) is approximately 0.530 \({Å}\). The radius for the first excited state (\(n = 2\)) 
orbit is (in \({Å}\)): 
 

• A. 0.13
• B. 1.06
• C. 4.77
• D. 2.12

Hint:

The radius of the Bohr orbit increases with the square of the principal quantum number, so the radius for the first excited state is four times the radius for the ground state.

Answer 1

The Correct Option is D

The radius of the Bohr orbit for any state \( n \) is given by the formula: \[ r_n = n^2 r_1 \] where: - \( r_n \) is the radius of the orbit for the \( n^{th} \) orbit, - \( r_1 \) is the radius of the Bohr orbit for \( n = 1 \), - \( n \) is the principal quantum number. Given that the radius for the ground state (\( n = 1 \)) is \( r_1 = 0.530 \, {Å} \), we can calculate the radius for the first excited state (\( n = 2 \)): \[ r_2 = 2^2 \times 0.530 = 4 \times 0.530 = 2.12 \, {Å} \] Thus, the radius for the first excited state (\( n = 2 \)) is \( 2.12 \, {Å} \).