Calculation of wavelength of electron in ground state:Radius of ground state of hydrogen atom = $0.53 \text{Å} = 0.53 \times 10^{-10} \mathrm{m}$According to de Broglie relation $2 \pi r = n \lambda$For ground state $\;\; n = 1$ $2 \times 3.14 \times 0.53 \times 10^{-10} \; = 1 \times \lambda$ $\therefore \;\; \lambda \; = 3.32 \times 10^{-10} \mathrm{m} \;$ $\; = 3.32 \text{\AA} \;$AlternativelyVelocity of electron, in the ground state, of hydrogen atom $= 2.18 \times 10^{-6} \mathrm{m} / \mathrm{s}$Hence momentum of revolving electron $p \; = m v$ $\; = 9.1 \times 10^{-31} \times 2.18$ $\times 10^{-6} \mathrm{kg} \mathrm{m} / \mathrm{s}$ $\lambda \; = \; \frac{h}{p}$ $= \; \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.18 \times 10^6} \mathrm{m}$ $\; = 3.32 \text{\AA}$ [Note : Also accept the following answer: Let $\lambda_n$ be the wavelength of the electron in the $n^{\text{th}}$ orbit, we then have $2 \pi r_n = n \lambda$ For ground state $n = 1$ $2 \pi r_n = \lambda$ ( $r = r_0$ is the radius of the ground state)[Alternatively$\lambda_n = \; \frac{h}{m v_n}$and $v_n = v_0$ (velocity of electron in ground state)$\lambda = \; \frac{h}{m v_n}$