Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 × 10^5 m/s and 2.5 × 10^5 m/s respectively. If the radii of these orbits are nearly same then the possible order of energy states are ____

Two electrons moving in hydrogen like atoms – Energy states from speed | JEE Main

Q. Two electrons are moving in orbits of two hydrogen like atoms with speeds \(3 \times 10^5\ \text{m/s}\) and \(2.5 \times 10^5\ \text{m/s}\) respectively. If the radii of these orbits are nearly same then the possible order of energy states are _____ respectively.

(A) 8 and 10

(B) 10 and 12

(D) 6 and 5

Correct Answer: 6 and 5

Step-by-Step Solution

According to Bohr model for a hydrogen like atom,

\[ v_n = \frac{Z e^2}{2\varepsilon_0 h} \cdot \frac{1}{n} \]

This shows that electron speed is inversely proportional to principal quantum number.

\[ v \propto \frac{1}{n} \]

Given that the radii of both orbits are nearly same, the nuclear charge factor cancels out and we can write

\[ \frac{v_1}{v_2} = \frac{n_2}{n_1} \]

Substitute the given values:

\[ \frac{3 \times 10^5}{2.5 \times 10^5} = \frac{n_2}{n_1} \]

\[ \frac{3}{2.5} = \frac{6}{5} \]

Thus, the possible quantum numbers are

\[ n_1 = 6,\quad n_2 = 5 \]

Therefore, the correct order of energy states is \[ \boxed{6 \text{ and } 5} \]

Step-by-Step Solution

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