Question

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

Answer 1

Answer:

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d

d

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Answer 2

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