Question

In H-atom, if ‘x’ is the radius of the first
Bohr orbit, de Broglie wavelength of an
electron in 3rd orbit is :

Answer 1

Answer:

rn=n^2 r1  

r3=9r1=9x

mvr=h/2π

mv9x=3 *h /2π

h/mv =6πx

λ=6πx

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Answer 2

Answer:

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Explanation:

For a hydrogen atom, radius of nth orbit, rn = n2h2/ 4π2mZe2 ,

  Therefore, r1/r3 = 12/32,

  r3 = 9/r1,

 = 9/x.

Now, according to De Broglie; angular momentum of electron in 3rd orbit is:

mvr3 = 3h/2π   or  h/mv = 2πr3/3,

And,

λ = h/mv   where, λ is the de Broglie wavelength.

Therefore,

λ = 2πr3/3,

 = 2π x 9/3x,

 = 6πx.

Hope it helps you.//✔✔✔

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