When an electron in a hydrogen atom jumps from the third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes

Step-by-step solution

The De Broglie wavelength is = h p , where h= Planck's constant and p= momentum of the electron. Also, momentum p=m v and kinetic energy E_K= 1 2 m v^2 From the above equation, we can relate momentum p and kinetic energy E_K as, p= 2 m E_K Now the ratio of wavelength of hydrogen atom when the atom jumps from third excited state to ground state viz (n=4 to n=1 ) ₁ ₂ = h p₁ h p₂ = p₂ p₁ = E_ K 2 E_ K 1 Also, we know E_ K n = -13.6 Z^2 n^2 Using equation (1), ₁ ₂ = 1^2 4^2 So, the ratio of the wavelength of ground sta