Q. 83.7( 6 Votes )

Answer :

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 = V_{0}/n

Where V is the velocity of the electron in any orbit , n is the state in which electron is orbiting,V_{0} is velocity of electron in first orbit of hydrogen atom

V_{0} = 2.18 × 10^{6} ms^{-1}

So velocity of electron in n = 2 state is

V = V_{0}/2 = 2.18 × 10^{6}ms^{-1}/2 = 1.09 × 10^{6} 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^{-34}Js

mass of an electron is

m = 9.1 × 10^{-31}Kg

The velocity of the electron is

v = 1.09 × 10^{6} 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

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