De-Broglie wavelength of an electron orbiting in the \( n = 2 \) state of hydrogen atom is close to (Given Bohr radius = \( 0.052\text{ nm} \))

For an electron in the \( n^{\text{th}} \) Bohr orbit, the circumference equals \( n\lambda \). Thus, \( \lambda = \frac{2\pi r_n}{n} \). For \( n = 2 \), \( r_2 = 2^2 \times r_0 = 4 \times 0.052\text{ nm} \), giving \( \lambda = 2\pi (0.052) \approx 0.67\text{ nm} \).