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

Question

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

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Solution

To find the frequency of the de-Broglie wave of an electron in Bohr's first orbit of a hydrogen atom, we first need to find the velocity of the electron in the first orbit.

The velocity (v) of an electron in the n-th orbit of a hydrogen atom is given by the formula:

v = 2πRH/nh

where RH is the Rydberg constant, n is the orbit number, and h is Planck's constant.

Substituting the given values:

v = 2π(2.18×10^-18 J)/(1)(6.6×10^-34 J.s) v = 2.07 x 10^6 m/s

The de-Broglie wavelength (λ) of a particle is given by the formula:

λ = h/mv

where m is the mass of the electron (9.1 x 10^-31 kg), v is the velocity of the electron, and h is Planck's constant.

Substituting the values:

λ = (6.6×10^-34 J.s)/(9.1 x 10^-31 kg)(2.07 x 10^6 m/s) λ = 3.37 x 10^-10 m

The frequency (f) of a wave is given by the formula:

f = v/λ

Substituting the values:

f = (2.07 x 10^6 m/s)/(3.37 x 10^-10 m) f = 6.14 x 10^15 Hz

So, the frequency of the de-Broglie wave of an electron in Bohr's first orbit of a hydrogen atom is approximately 6 x 10^15 Hz.

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