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The de-Broglie wavelength of an electron in the first Bohr orbit is

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equal to twice the circumference of first orbit

equal to half the circumference of first orbit

equal to the circumference of the first orbit

equal to one-fourth the circumference of the first orbit

answer is D.

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Detailed Solution

According to Bohr the circumference of the orbital will be integral multiple of de-Broglie wavelength.

$\Rightarrow nλ = 2 {πr}_{n}$

or first orbit n=1 so λ=2πr₁.

$λ$ equals to the circumference of the first orbit.

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