An electron in a hydrogen atom has a de Broglie wavelength ₀ in the ground state. In a particular excited state, its de Broglie wavelength is 3 ₀ . The ratio of the area of the electron's orbit in this excited state to the area of its ground state orbit is

Step-by-step solution

From Bohr's quantization condition, mvr = nh 2 . The de Broglie wavelength is = h mv . Substituting mv from the first equation, we get = 2 r n . For a hydrogen atom, the radius of the n^ th orbit is r n^2 . Therefore, n^2 n n . Given that the de Broglie wavelength in the excited state is 3 ₀ , the principal quantum number of this state is n = 3 . The area of the circular orbit is A = r^2 . Since r n^2 , we have A (n^2)^2 = n^4 . The ratio of the area of the orbit in the excited state ( n=3 ) to that in the ground s