Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

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.

Answer 1

Concept Used:

According to de Broglie, a moving electron behaves like a wave with wavelength:

λ = h / mv

where h is Planck’s constant, m is mass of electron, and v is its velocity.

Step 1: Expression for circumference of Bohr orbit

For an electron revolving in the n^th Bohr orbit of radius r_n,

Circumference = 2πr_n

Step 2: Bohr’s quantization condition

According to Bohr,

mvr_n = n (h / 2π)

Multiplying both sides by 2π:

2πmvr_n = nh

Step 3: Rearrangement

2πr_n = n (h / mv)

But,

h / mv = λ (de Broglie wavelength)

Step 4: Final relation

2πr_n = nλ

Conclusion:

The circumference of the Bohr orbit is an integral multiple of the de Broglie wavelength associated with the electron.

Final Answer:

2πr_n = nλ

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Solution and Explanation

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