Q113 / 150 · MHT CET 2026 (13 April Shift 1) · Physics
The ratio of centripetal acceleration for an electron revolving in 3rd and 5th Bohr orbit of hydrogen atom is
- A$25 : 9$
- B$125 : 27$
- Correct answerC$625 : 81$
- D$5 : 3$
Solution
Centripetal acceleration is given by $a_c = \dfrac{v^2}{r}$
In Bohr's model of hydrogen atom, velocity $v \propto \dfrac{1}{n}$ and radius $r \propto n^2$
Substituting these proportionalities, we get $a_c \propto \dfrac{(1/n)^2}{n^2} \Rightarrow a_c \propto \dfrac{1}{n^4}$
For the 3rd and 5th orbits, the ratio of centripetal accelerations is:
$\dfrac{a_3}{a_5} = \left(\dfrac{5}{3}\right)^4 = \dfrac{625}{81}$
In Bohr's model of hydrogen atom, velocity $v \propto \dfrac{1}{n}$ and radius $r \propto n^2$
Substituting these proportionalities, we get $a_c \propto \dfrac{(1/n)^2}{n^2} \Rightarrow a_c \propto \dfrac{1}{n^4}$
For the 3rd and 5th orbits, the ratio of centripetal accelerations is:
$\dfrac{a_3}{a_5} = \left(\dfrac{5}{3}\right)^4 = \dfrac{625}{81}$
Hence, the correct answer is option: 'C'.