Calculate the de-Broglie wavelength of the electron orbitting in the n = 2 state of hydrogen atom.

We know in an hydrogen atom an negatively charged electron is revolving around nucleus of hydrogen consisting of one proton, orbit is the well-defined circular path in which electron revolves around the nucleus As Shown in figure

these orbits have the different energy level, and electron have different speed depending upon the orbit in which it is revolving, the movement of an electron can be characterized by the state such as n = 1 for orbiting in the first orbit, n = 2 for orbiting in the second orbit

The velocity of electron also depends on its state and is given as

V = V₀/n

Where V is the velocity of the electron in any orbit , n is the state in which electron is orbiting,V₀ is velocity of electron in first orbit of hydrogen atom

V₀ = 2.18 × 10⁶ ms^-1

So velocity of electron in n = 2 state is

V = V₀/2 = 2.18 × 10⁶ms^-1/2 = 1.09 × 10⁶ ms^-1

Now we know de Broglie wavelength of a Particle is given by relation