Question

In a H-like ion, ratio of speed of electron in two orbit is 3 : 2, then ratio of energies in these orbits should be :

• A. $\frac{3}{5}$
• B. $\frac{9}{4}$
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Hint:

You don't need to find the quantum numbers $n_1$ and $n_2$. Since Kinetic Energy $K = \frac{1}{2}mv^2$ and Total Energy $E = -K$, the ratio of total energies is exactly the same as the ratio of their kinetic energies. Hence, $E_1/E_2 = v_1^2 / v_2^2$ directly.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are dealing with a hydrogen-like ion. We are given the ratio of the speeds of the electron in two different orbits and asked to find the ratio of their total energies in these orbits.
Step 2: Key Formula or Approach:
From Bohr's model of the atom:
1. The speed of an electron in the $n^{th}$ orbit is directly proportional to $Z/n$. So, $v \propto \frac{Z}{n}$. For the same ion, Z is constant, so $v \propto \frac{1}{n}$.
2. The total energy of an electron in the $n^{th}$ orbit is proportional to $Z^2/n^2$. So, $E \propto -\frac{Z^2}{n^2}$. The magnitude of energy is $|E| \propto \frac{1}{n^2}$.
From these relations, we can see that Energy magnitude is proportional to the square of the velocity ($E \propto v^2$).
Step 3: Detailed Explanation:
Let the two orbits be $n_1$ and $n_2$.
The ratio of speeds is given as:
$\frac{v_{1}}{v_{2}} = \frac{3}{2}$
Since $v \propto \frac{1}{n}$, we have:
$\frac{n_{2}}{n_{1}} = \frac{v_{1}}{v_{2}} = \frac{3}{2}$
Now, the total energy of the electron is $E = -13.6 \frac{Z^2}{n^2}$ eV.
The ratio of energies is:
$\frac{E_{1}}{E_{2}} = \frac{-13.6 \frac{Z^2}{n_{1}^2}}{-13.6 \frac{Z^2}{n_{2}^2}} = \frac{n_{2}^2}{n_{1}^2}$
Substitute the value of $n_{2}/n_{1}$:
$\frac{E_{1}}{E_{2}} = \left(\frac{n_{2}}{n_{1}}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$
Note: The question asks for the ratio of energies. Total energy is negative, but ratios of the energy states or their magnitudes are positive.
Step 4: Final Answer:
The ratio of energies in these orbits is $\frac{9}{4}$.