Question

Two electrons are moving in orbits of two hydrogen like atoms with speeds $3\times10^{5}\,\text{m/s}$ and $2.5\times10^{5}\,\text{m/s}$ respectively. If the radii of these orbits are nearly same then the possible order of energy states are ___ respectively.

• A. $10$ and $12$
• B. $8$ and $10$
• C. $6$ and $5$
• D. $9$ and $8$

Hint:

In hydrogen like atoms, electron speed is inversely proportional to the principal quantum number.

Answer 1

The Correct Option is C

Step 1: Using Bohr model relation for speed.
For a hydrogen like atom, the speed of electron in the $n^{\text{th}}$ orbit is given by:
\[ v_n \propto \dfrac{1}{n} \] Step 2: Writing ratio of speeds.
\[ \dfrac{v_1}{v_2} = \dfrac{3\times10^{5}}{2.5\times10^{5}} = \dfrac{6}{5} \] Step 3: Relating speed ratio to principal quantum number.
\[ \dfrac{v_1}{v_2} = \dfrac{n_2}{n_1} \] \[ \dfrac{n_2}{n_1} = \dfrac{6}{5} \] Step 4: Determining energy states.
Thus, the possible values of $n_1$ and $n_2$ are $6$ and $5$ respectively.
Step 5: Final conclusion.
The correct order of energy states is $6$ and $5$.