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According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is 10-9. The principle quantum number for this electron is

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a

1

b

2

c

3

d

4

answer is C.

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Detailed Solution

mvrn=nh2π, 2πrn=nλ, 2π(0.53X10-10)n2Z=nλ(z=1), n=λ2πX0.53X10-10=10-92×3.14×0.53×10-10=3

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