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.
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Answered by
17
Heya
here's your answer...
Velocity of electron = 1275 × Velocity of light
=1275× 3.0 × 1010
=1.09×108 cm/s
Now Un=2π e2nh
n= 2π e2un×h
=2×3.14×(4.803×10−10)21.09×108×6.625×10−27
=2.006
=2
hope it helps...✌✌
here's your answer...
Velocity of electron = 1275 × Velocity of light
=1275× 3.0 × 1010
=1.09×108 cm/s
Now Un=2π e2nh
n= 2π e2un×h
=2×3.14×(4.803×10−10)21.09×108×6.625×10−27
=2.006
=2
hope it helps...✌✌
Answered by
0
Answer:
here we use Vn1/Vn2=n2/N1
By solving we get n=2
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