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Question

The de-Broglie wavelength of an electron in first orbit of Bohr's hydrogen is equal to

A
radius of the orbit
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B
perimeter of the orbit
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C
diameter of the orbit
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D
half of the perimeter of the orbit
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Solution

The correct option is B perimeter of the orbit
According to the expression given by de-Broglie for the wavelength of a particle of momentum mv
λ=hmv -------(1)
According to the Bohr's postulate, electrons move only in those orbits where angular momentum is quantised.
mvr=nh2π ---------(2)
Putting the expression for mv from equation 1 to equation 2,
we get nλ=2πr
For first orbit n=1
Hence λ=2πr
which is perimeter of the orbit.

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