Question

the circumference of the second orbit of hydrogen atom if the wavelength of the electron is 5 into 10 to the power minus 9​

Answer 1

Answer:

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 : The circumference of second orbit of H-atom is, $10^{-8}m$

Explanation : Given,

Wavelength of electron = $5\times 10^{-9}m$

The Bohr's formula is :

$2\pi r=n\lambda$

where,

r = radius of hydrogen atom

n = number of orbit = 2

$\lambda$ = wavelength of electron

$2\pi r$ = circumference of hydrogen atom

Now put all the given values in the above formula, we get the circumference of hydrogen atom.

$2\pi r=2\times (5\times 10^{-9}m)=1\times 10^{-8}m$

Therefore, the circumference of second orbit of H-atom is, $10^{-8}m$