Question

Radius of first orbit in hydrogen atom is x 2 A 0 . Wavelength of electron moving in 4th orbit of hydrogen atom in angstroms will be

Answer 1

Answer:

de-Broglie wavelength ,λ=hp ....(i)

According to Bohr's quantisation condition,

mvrn=nh2π⇒prn=nh2π ....(ii)

From Eqs. (i) and (ii) we get

λ=h×2πrnnh⇒λ=2πrnn

For fourth orbit (n=4)⇒λ=2πr44 ....(iii)

Moreover,r∝n2⇒r1r4≡(1)2(4)2⇒r4=16r

Subtituting the value of r4 in Eq. (iii), we get

λ=2π|16r1|4=8πr1=8πr

Answer 2

$\Huge \tt8\pi \R$