Question

28. The circumference of second orbit of H-atom, if the
wavelength of electron is 5 x 10 m
(1) 50 × 10 m (2) 10 m
(3) 10-10m
(4) 10-18m​

Answer 1

n¥ = 2πr

where r is radius of orbit

so r = 0.53 n²/z

n is orbit ( here n =2)

and ¥ is lemda i.e wavelength

substitiing all the values

u ll get the value of r

and circumference is equal 2πr

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Answer 2

Answer:

(3)-10^-10m

Explanation:

n¥=2Πr

n=no. of orbit, ¥=lambda (wavelength)

Circumference=2Πr

Substituting values

2×5×10^-9=10^-10m

So 10^-10=2Πr