The revolution per second for the electron present in 3rd orbit of He+ ion is
9.73 × 1014
6.57 × 1015
9.73 × 1016
1.52 × 1016
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The number of revolutions per second is 2 x 10^16
Option (D) is correct.
Explanation:
Angular frequency is given by:
mvr = nh / 2π
mωr^2 = nh / 2π
Orbital frequency (u) = nh / 4π^2 mv^2
Now put the values in the above formula.
Orbital frequency (u) = 3 x 6.67 x 10^-34 / 4 x (3.14)^2 x (9.1 x 10^-31) x (0.53 x 10^-10)^2
Orbital frequency (u) = 20.01 x 10^-34 / 39.44 x (9.1 x 10^-31) x 0.28 x 10-20
Orbital frequency (u) = 20.01 x 10^-34 / 100 x 10^-31-20
= 0.2 x 10^-34+51
Orbital frequency (u) = 2 x 10^16
Thus the number of revolutions per second is 2 x 10^16
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