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Question

In H-atom, if ‘x’ is the radius of first Bohr orbit, de Broglie wavelength of an electron in 3rd orbit is:


A

3πx

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B

6πx

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C

9x2

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D

x2

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Solution

The correct option is B

6πx


rn=0529×n2ZA

r1=0.529×(1)2(1)=x

r3=0.529×(3)2(1)=0.529×91=9x

We know that,

2πr=nλ2π9x=(3)λλ=6πx


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