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:
The question asks for the ratio of the kinetic energy ($K.E.$) to the total mechanical energy ($E$) of an electron orbiting within the $n^{\text{th}}$ stable energy shell of a hydrogenic atom according to the Bohr atomic model.

Step 2: Key Formula or Approach:
According to the electrostatic physics of a bound system obeying Coulomb's law, the energy terms share a fixed relationship: $$\text{Kinetic Energy } (K.E.) = \frac{kZe^2}{2r}$$ $$\text{Potential Energy } (P.E.) = -\frac{kZe^2}{r}$$ $$\text{Total Energy } (E) = K.E. + P.E. = -\frac{kZe^2}{2r}$$ This gives the clean proportional identity: $E = -K.E.$

Step 3: Detailed Explanation:
From the relations above, we can directly link the values of kinetic energy and total energy: $$E = -K.E. \implies \frac{K.E.}{E} = \frac{K.E.}{-K.E.} = -1$$ Writing this relationship as a ratio of kinetic energy to total energy yields: $$\text{Ratio} = 1 : -1$$ The negative sign indicates that the total energy is negative, which physically means the electron is trapped inside a bound potential well.

Step 4: Final Answer:
The ratio of kinetic energy to total energy is $1 : -1$, which corresponds to option (B).