When an electron in a hydrogen atom jumps from the excited state to the ground state, how would the de Broglie wavelength associated with the electron change?

Kinetic energy of an electron in an orbit in Bohr's model for a Hydrogen atom is given by: E = 13.6/n² eV

de Broglie wavelength of an electron is given by:
λ = h/p

But, p = √(2mE)

λ = h/√(2mE) => λ ∝ 1/√E

Hence, when an electron jumps from the excited state to the ground state, its kinetic energy (E) increases, and therefore its de Broglie wavelength (λ) decreases.