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

$r=0.529\times \frac{n^2}{z}$
$r\propto n^2$
$R\propto (2{)}^2$
$R_1\propto (3{)}^2$
Divide (1) by (2) $\frac{R_1}{R}=\frac{4}{9}$
$R_1=\frac{9R}{4}$
$R_1=2.25R$

Answer 2

Concept:

• Bohr postulate 1: angular momentum is quantized, $mvr = \dfrac{nh}{2\pi}$.
• Bohr postulate 2: the electrostatic force of attraction supplies the centripetal force needed for circular motion, $\dfrac{mv^2}{r} = \dfrac{kZe^2}{r^2}$.
• Combining these two equations derives the radius formula directly, instead of just quoting it.

Step 1: Write the velocity from the angular momentum condition.
$mvr = \dfrac{nh}{2\pi} \Rightarrow v = \dfrac{nh}{2\pi mr}$

Step 2: Substitute this velocity into the force balance equation.
$\dfrac{mv^2}{r} = \dfrac{kZe^2}{r^2}$
$\dfrac{m}{r}\left(\dfrac{nh}{2\pi mr}\right)^2 = \dfrac{kZe^2}{r^2}$

Step 3: Simplify to isolate $r$.
$\dfrac{n^2h^2}{4\pi^2mr^3} = \dfrac{kZe^2}{r^2}$
$r = \dfrac{n^2h^2}{4\pi^2 m k Z e^2}$
This shows $r \propto \dfrac{n^2}{Z}$.

Step 4: Apply the proportionality to the given orbits.
Since both orbits belong to the same atom, $Z$ is the same for both, so it cancels in the ratio and $r \propto n^2$ alone.
$\dfrac{r_3}{r_2} = \dfrac{3^2}{2^2} = \dfrac{9}{4}$

Step 5: Solve for the third orbit radius.
$r_2 = R$ (given)
$r_3 = \dfrac{9}{4}R = 2.25R$

Final Answer: $2.25R$