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Question

If the radius of first Bohr orbit is x, then de-Broglie wavelength of electron in 3rd orbit is nearly:


A

2πx

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B

6πx

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C

9x

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D

x3

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Solution

The correct option is B

6πx


rn=n2r1

So, the third radius would be r3=9r1=9x

angular momentum is given by: mvr=h/2π

mv(9x)=3h/2π

h/mv=6πx

λ=6πx


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