What is the ratio of the de Broglie wavelength of an electron revolving in the third orbit of a singly ionized helium atom ( He^+ ) to that of an electron revolving in the second orbit of a doubly ionized lithium atom ( Li²⁺ )?

Step-by-step solution

According to Bohr's quantization rule, the angular momentum is mvr = nh 2 . The de Broglie wavelength of an electron is given by = h mv . Substituting mv from the first equation, we get = 2 r n . The radius of the n -th orbit of a hydrogen-like atom is r n^2 Z . Therefore, the de Broglie wavelength is proportional to n Z . For the electron in the third orbit of He^+ , n₁ = 3 and Z₁ = 2 . ₁ 3 2 For the electron in the second orbit of Li²⁺ , n₂ = 2 and Z₂ = 3 . ₂ 2 3 The ratio of their de Broglie wavelengths is ₁ ₂ =