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Q.

The radius of the 4th Bohr's orbit is 0.864 nm. The de Broglie wavelength of the electron in that orbit is___

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a

13.297 Ao

b

1.3565 nm

c

1.3291 pm

d

0.1329 nm

answer is B.

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Detailed Solution

2πrn=nλ⇒ λ=2πrnn=2×3.14×0.8644nm=1.3565nm
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