Question

The radius of electron's second stationary orbit in Bohr's atom is $R$ The radius of 3rd orbit will be

• A. $\frac{ R }{3}$
• B. $9 R$
• C. $2.25 R$
• D. $3 R$

Hint:

Derive the radius formula from the two Bohr postulates, quantized angular momentum and centripetal force balance, instead of recalling it directly. This shows exactly why radius depends only on $n^2$ for a fixed atom, which is all you need for the ratio.

Answer 1

The Correct Option is C

To solve this problem, we need to understand the Bohr model of the hydrogen atom, which explains the electron's energy levels and their corresponding radii in an atom.

In the Bohr model, the radius of an electron's orbit is given by the formula:

\(r_n = n^2 \times r_1\)

where:

• \(r_n\) is the radius of the nth orbit,
• \(n\) is the principal quantum number (orbit number),
• \(r_1\) is the radius of the first orbit, which is Bohr's radius.

For the second orbit (n = 2), the radius is given as \(R\). Thus, we can write:

\(R = 2^2 \times r_1\)

Thus, substituting the known values, we have:

\(R = 4r_1\)

Now, let's find the radius of the third orbit (n = 3):

\(r_3 = 3^2 \times r_1 = 9r_1\)

To find the relationship between the second orbit radius \(R\) and the third orbit radius \(r_3\), we divide \(r_3\) by \(R\):

\(\frac{r_3}{R} = \frac{9r_1}{4r_1} = \frac{9}{4} = 2.25\)

Therefore, the radius of the third orbit compared to the second orbit is \(2.25R\).

Thus, the correct answer is \(2.25R\).