Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atom is _________ × 10¹³ Hz (nearest integer).
[Given : R_H (Rydberg constant) =2.18 × 10⁻¹⁸ J, h (Plank's constant) =6.6 × 10⁻³⁴ J . s.]

Solution

Chemistry ma’am · 4 steps

Frequency is speed over wavelength, and with λ = h/mv that becomes 2 × KE/h.

1. Frequency of the wave. For any wave, ν = v/λ. The de Broglie wavelength of the electron is λ = h/mv, so ν = mv²/h.

   ν = mv²/h = (2 × KE)/h

2. Kinetic energy. In a Bohr orbit, KE = |Eₙ|. For the first orbit of hydrogen, |E₁| = R_H = 2.18 × 10⁻¹⁸ J, so 2 × KE = 4.36 × 10⁻¹⁸ J.
3. Put in numbers. With the precise value h = 6.626 × 10⁻³⁴ J s, ν = (4.36 × 10⁻¹⁸)/(6.626 × 10⁻³⁴) = 6.58 × 10¹⁵ Hz. The answer key uses this value. With the rounded h = 6.6 × 10⁻³⁴ given in the question, the same working gives about 6.61 × 10¹⁵ Hz.
4. So. ν = 658 × 10¹³ Hz, so the answer is 658.

The common slip: 330. Using ν = E/h and dropping the factor 2 gives about 330, which is exactly half the right value.

The idea to keep. For de Broglie frequency, use ν = v/λ and then λ = h/mv. That gives ν = 2 KE/h, not KE/h. In a Bohr orbit, KE equals the magnitude of the total energy.

Answer from the NTA final answer key. Solution by PrepWiser.