The radius of the third stationary orbit of an electron in Bohr's atom is \( R \). The radius of the fourth stationary orbit will be:
- \(\frac{4}{3} R\)
- \(\frac{16}{9} R\)
- \(\frac{3}{4} R\)
- \(\frac{9}{16} R\)
The Correct Option is B
Approach Solution - 1
To determine the radius of the fourth stationary orbit relative to the third orbit in Bohr's atom model, we need to understand the relation between the orbit radius and the principal quantum number.
According to Bohr's model, the radius \(r_n\) of the \(n\)-th orbit is given by the formula:
\(r_n = n^2 \cdot r_1\)
where \(r_1\) is the radius of the first stationary orbit, and \(n\) is the principal quantum number.
Given:
The radius of the third orbit is \(R\). Therefore, we have:
\(r_3 = 3^2 \cdot r_1 = 9 \cdot r_1 = R\)
From this, we can find \(r_1\):
\(r_1 = \frac{R}{9}\)
We want to find the radius of the fourth orbit, \(r_4\):
\(r_4 = 4^2 \cdot r_1 = 16 \cdot r_1\)
Substituting the value of \(r_1\), we get:
\(r_4 = 16 \cdot \frac{R}{9} = \frac{16}{9} R\)
Thus, the radius of the fourth stationary orbit is \(\frac{16}{9} R\).
Therefore, the correct answer is: \(\frac{16}{9} R\).
This matches the given correct answer option, confirming our solution is accurate.
Approach Solution -2
According to Bohr’s theory, the radius of an electron’s orbit is given by:
\[ r \propto \frac{n^2}{Z} \]
where \(n\) is the principal quantum number and \(Z\) is the atomic number. Since the electron is in hydrogen (\(Z = 1\)), we get:
\[ \frac{r_4}{r_3} = \frac{4^2}{3^2} = \frac{16}{9} \]
Thus, \(r_4 = \frac{16}{9}R\).
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