Question

In $n^{\text{th}}$ Bohr orbit, the ratio of the kinetic energy of an electron to the total energy of it, is

• A. $2 : 1$
• B. $1 : -1$
• C. $+1 : 1$
• D. $-1 : 2$

Hint:

To remember this quickly, use the absolute value rule for virial systems: $\text{Kinetic Energy} = |\text{Total Energy}|$. Since kinetic energy must always be positive and total energy for any stable orbit must always be negative, their ratio must be a negative value, instantly pointing to choice (B) or (D).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Find the ratio of kinetic energy to total energy for an electron in the nth Bohr orbit of a hydrogen-like atom.

Step 2: Key Formula or Approach:
In a Coulomb bound system: K.E. = kZe²/(2r), P.E. = –kZe²/r, Total E = –kZe²/(2r) = –K.E.

Step 3: Detailed Explanation:
K.E. / E = K.E. / (–K.E.) = –1. The ratio is 1 : –1.

Step 4: Final Answer:
Ratio is 1 : –1, matching option (B).