The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10-16 s. The frequency of revolution of the electron in its first excited state (in s-1) is: (a) 5.6 × 1012 (b) 1.6 × 1014 (c) 7.8 × 1014 (d) 6.2 × 1015

Answer: (c). T ∝ n³, so T₂ = 8T₁ = 8×1.6×10⁻¹⁶ = 1.28×10⁻¹⁵ s f₂ = 1/T₂ = 11.28×10⁻¹⁵ ≈ 7.8×10¹⁴ s⁻¹ So the answer is (c) 7.8×10¹⁴.

The time period of revolution of electron in its ground state orbit… | JEE Main 2020 (7 Jan, Shift 1)

Q98JEE Main 2020 · 7 JAN · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum

The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 $\times$ 10-16 s. The frequency of revolution of the electron in its first excited state (in s-1) is:

Show answer and solution

Answer: (c)

$T \propto n^3$ , so $T_2 = 8T_1 = 8\times1.6\times10^{-16} = 1.28\times10^{-15}\text{ s}$
$f_2 = \frac{1}{T_2} = \frac{1}{1.28\times10^{-15}} \approx 7.8\times10^{14}\text{ s}^{-1}$
So the answer is (c) $7.8\times10^{14}$ .

Report a mistake

Bohr Model: Orbits, Speed & Angular Momentum question from JEE Main 2020 (7 Jan, Shift 1)

More questions on Bohr Model: Orbits, Speed & Angular Momentum