Question:medium

Ratio of centripetal acceleration for an electron revolving in 3rd and 5th Bohr orbit of hydrogen atom is

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Instead of memorizing the full expression for acceleration, just remember the fundamental scaling factors: radius grows as $n^2$ and velocity drops as $1/n$. Combining them always shows that any acceleration parameters scale extremely quickly as an inverse fourth power ($1/n^4$).
Updated On: Jun 4, 2026
  • $425 : 18$
  • $625 : 81$
  • $125 : 27$
  • $221 : 36$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: What we want.
An electron goes round the nucleus in circles called Bohr orbits. We compare how strongly it is pulled inward (its centripetal acceleration) in the 3rd orbit versus the 5th orbit.
Step 2: Meaning of centripetal acceleration.
For any object moving in a circle, the inward acceleration is \[ a = \frac{v^2}{r} \] where $v$ is its speed and $r$ is the radius of its circle.
Step 3: How speed and radius depend on the orbit.
In Bohr's model the speed gets smaller for higher orbits and the radius gets bigger: \[ v \propto \frac{1}{n} \quad\text{and}\quad r \propto n^2 \] Here $n$ is the orbit number.
Step 4: Build the rule for acceleration.
Put these into $a = v^2/r$: \[ a \propto \frac{(1/n)^2}{n^2} = \frac{1}{n^4} \] So acceleration falls very fast as $n$ grows. It is smaller for higher orbits.
Step 5: Make the ratio.
Since $a \propto 1/n^4$, the orbit with the smaller $n$ has the larger $a$: \[ \frac{a_3}{a_5} = \frac{1/3^4}{1/5^4} = \left(\frac{5}{3}\right)^4 \]
Step 6: Work out the numbers.
$5^4 = 625$ and $3^4 = 81$. So \[ \frac{a_3}{a_5} = \frac{625}{81} \] The ratio is $625 : 81$, which is option (2). \[ \boxed{a_3 : a_5 = 625 : 81} \]
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