Question

find out the wavelength of the electron orbiting in the ground state of hydrogen atom​

Answer 1

Explanation:

Radius of ground state of hydrogen atom

=0.53 \dot{A} = 0.53×10 ^{-10} m=0.53A˙ =0.53×10−10m

According to de Brogile relation

2\pi r = n\lambda2πr=nλ

For ground state n = 1n=1

2×3.14×0.53×10^{-10}=1×\lambda2×3.14×0.53×10−10=1×λ

\therefore \lambda =3.32×10 ^{-10}m∴ λ=3.32×10−10m

=3.32\dot{A} =3.32A˙