Calculation of wavelength of electron in ground state:




Radius of ground state of hydrogen atom = 0.53Å=0.53×10−10m
According to de Broglie relation 2πr=nλ
For ground state ‌‌n=1
2×3.14×0.53×10−10‌=1×λ
∴‌‌λ‌=3.32×10−10m‌
‌=3.32Å‌
Alternatively
Velocity of electron, in the ground state, of hydrogen atom
=2.18×10−6m∕ s
Hence momentum of revolving electron
p‌=mv
‌=9.1×10−31×2.18 ×10−6‌kg‌m∕ s
λ‌=‌

h

p

=‌

6.63×10−34

9.1×10−31×2.18×106

m
‌=3.32Å
[Note : Also accept the following answer:
Let λn be the wavelength of the electron in the n‌th ‌ orbit, we then have
2πrn=nλ
For ground state n=1
2πrn=λ
( r=r0 is the radius of the ground state)
[Alternatively
λn=‌

h

mvn

and vn=v0 (velocity of electron in ground state)
λ=‌

h

mvn