Question

show that the velocity un Bohrs orbit is inversely proportional to the principal quantum number​

Answer 1

Answer:

Show that angular velocity of electron Bohr orbit is inversely proportional to cube of principal quantum number, (4 marks Time or frequency of revolution f = Ans: Let m = mass of the electron, -e= charge on the electron, + = charge on proton, 1. ... = linear velocity of electron in orbit, n = principal quantum number, 2.

Answer 2

To Prove:

The Bohr velocity is inversely proportional to the principal quantum number (n).

Proof:

We all know that Neil Bohr contributed tremendously to the field of Quantum Physics with the help of his postulates.

The Bohr model of atom states that in an atom, electrons revolve around the positively charged nucleus in a definite circular path called orbits or shells and each orbit or shell has a fixed energy and these circular orbits are known as orbital shells.

Now, from Bohr's postulates we can say that the centripetal force required for an electron to revolve around the nucleus is provided by the electrostatic force of attraction between the electron and the nucleus;

i.e.,  $\frac{mv^{2} }{r}$ = $\frac{ze^{2} }{4\pi r^{2} }$ × 1/ε₀                         ----------(1)

where; z is the atomic number

            e is the electronic charge

            m is the mass of the electron

            r is the radius of orbit of the electron

            v is the velocity of  the electron

Also, the angular momentum of an electron is an integral multiple of  $\frac{h}{2\pi }$ , e h is Planck's constant.

i.e.,  $mvr = \frac{nh}{2\pi }$                                  -----------(2)

Now, By dividing equation (1) by (2), we get;

v  =  $\frac{ze^{2} }{4\pi }$× 1/ε₀ × $\frac{nh}{2\pi }$

∴ v = $\frac{ze^{2} }{2nh}$× 1/ε₀

Let, A = $\frac{ze^{2} }{2h}$× 1/ε₀

∴ v = $\frac{A}{n}$

∴ v ∝ $\frac{1}{n}$

From the above-derived expression of the velocity of an electron in an orbit, it is clear that the velocity is inversely proportional to the principal quantum number (n).

Hence Proved.

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