The de Broglie wavelength of an electron in a Bohr orbit of a hydrogen atom is 6 a₀ , where a₀ is the Bohr radius. If E₀ is the magnitude of the ground state energy of the hydrogen atom, the potential energy of the electron in this orbit is:

Step-by-step solution

According to de Broglie's hypothesis and Bohr's model, the circumference of the n^ th orbit is an integral multiple of the de Broglie wavelength : 2 r_n = n The radius of the n^ th Bohr orbit is given by: r_n = n^2 a₀ Substituting r_n into the first equation: 2 (n^2 a₀) = n = 2 n a₀ Given that = 6 a₀ , we have: 2 n a₀ = 6 a₀ n = 3 The total energy of the electron in the n^ th orbit is: E_n = - E₀ n^2 = - E₀ 9 In a hydrogen atom, the potential energy U is twice the total energy E : U_n = 2E_n = - 2E₀ 9 Answer: - 2E₀