# 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

𝜆 = h/mv

Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v

We know the value of Planck’s constant

h = 6.626 × 10-34Js

mass of an electron is

m = 9.1 × 10-31Kg

The velocity of the electron is

v = 1.09 × 106 ms-1

so putting these values to get the de-Broglie wavelength of the electron i.e the de-Broglie wavelength of the electron is

𝜆 = 0.668 nm

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.