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: Use centripetal force balance in Bohr model

In a hydrogen-like ion, the electrostatic force provides the required centripetal force for the electron:

m v² / r = k Z e² / r²

From this relation,

v² ∝ Z / r

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Step 2: Express total energy in terms of velocity

The total energy of the electron is:

E = K + U

Using Bohr model results,

E = − (1/2) m v²

Thus, the magnitude of total energy is directly proportional to v².

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Step 3: Use given velocity ratio

Given:

v₁ / v₂ = 3 / 2

Square both sides:

(v₁ / v₂)² = 9 / 4

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Step 4: Relate energy ratio to velocity ratio

Since |E| ∝ v²,

E₁ / E₂ = (v₁ / v₂)²

E₁ / E₂ = 9 / 4

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Final Answer:

The ratio of energies in the two orbits is
9 / 4