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The velocity of an electron in a certain Bohr's orbit of H-atom bears the ratio 1:275 to the velocity of light. Then find the quantum numbers (n) of orbit.

$(a)\;2\qquad(b)\;1\qquad(C)\;3\qquad(d)\;4$

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Velocity of electron = $\large\frac{1}{275}$ $\times$ Velocity of light
$=\large\frac{1}{275}$$\times$ 3.0 $\times$ $10^{10}$
$=1.09 \times 10^8$ cm/s
Now $U_n =\large\frac{2\pi\ e^2} {nh}$
n= $\large\frac{2\pi\ e^2}{u_n\times h}$
=$ \large\frac{2\times3.14\times(4.803\times10^{-10})^2}{1.09\times10^8\times6.625\times10^{-27}}$
=2.006
=2
Hence the answer is (a)
answered Jan 24, 2014 by sharmaaparna1
 

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