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

To determine the energy states of the two hydrogen-like atoms, we can use the energy formula for electrons in an orbit, which is specific to hydrogen-like atoms:

The energy of an electron in the \(n\)-th orbit of a hydrogen-like atom is given by: \(E_n = - \left(\frac{13.6\,\text{eV}}{n^2}\right) \frac{Z^2}{1}\)

Since we are given that the speeds of the electrons in their respective orbits are nearly constant and the radii of the orbits are nearly the same, we can relate the speed of the electron to its energy level.

• Using the formula for the kinetic energy of an electron in a circular orbit: \(K.E = \frac{1}{2}mv^2\)
• For a hydrogen-like atom, it follows that: \(v \propto \frac{Z}{n}\) (where \(v\) is the speed, \(Z\) is the atomic number which is 1 for hydrogen-like atoms, and \(n\) is the principal quantum number).
• Given \(v_1 = 3 \times 10^5 \text{ m/s}\) and \(v_2 = 2.5 \times 10^5 \text{ m/s}\), we have: \(\frac{v_1}{v_2} = \frac{n_2}{n_1}\).

Then: \(\frac{3 \times 10^5}{2.5 \times 10^5} = \frac{n_2}{n_1}\), leading to: \(\frac{3}{2.5} = \frac{n_2}{n_1}\)

Simplifying the fraction, we have: \(\frac{6}{5} = \frac{n_2}{n_1}\), which implies that: \(n_1 = 6\) and \(n_2 = 5\).

Thus, the possible order of energy states are 6 and 5, respectively.

The correct answer is:

$6$ and $5$