Question

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

• A. \( \frac{424}{21} \)
• B. \( \frac{625}{81} \)
• C. \( \frac{125}{4} \)
• D. \( \frac{775}{61} \)

Hint:

The centripetal acceleration in Bohr orbits is inversely proportional to the square of the orbit's radius, which is proportional to the square of the orbit number.

Answer 1

The Correct Option is B

Step 1: Formula for centripetal acceleration.
The centripetal acceleration \( a_c \) for an electron in the \( n \)-th orbit of a hydrogen atom is given by: \[ a_c = \frac{v^2}{r} = \frac{k e^2}{m e r^2} \] where \( v \) is the speed of the electron, \( r \) is the radius of the orbit, and \( k \) is the Coulomb's constant. The radius of the \( n \)-th orbit is proportional to \( n^2 \), i.e., \( r_n \propto n^2 \). Step 2: Calculating the ratio.
Thus, the ratio of the centripetal accelerations for the 3rd and 5th orbits is: \[ \frac{a_{c,3}}{a_{c,5}} = \left( \frac{5}{3} \right)^2 = \frac{625}{81} \] Step 3: Conclusion.
Thus, the correct answer is (B) \( \frac{625}{81} \).