Home | Courses | Back to Chapter | Any Question? |

PHYSICS-12

Chapter 20 - Atomic Models

Radius of nth stationery orbit for Hydrogen atom.


Bohr's assumed that hydrogen atom consist of unipositive charge concentrated on a nucleus and unit negative charge concentrated on a nucleus and unit negative charge.

The electrons which revolve around the nucleus is balanced by centripetal force as shown in above figure.
The electrostatic force of attraction between electron and nucleus is given by:

Fe= (1/4πε₀)(ze2/r2) <------- (i)
where, ε₀ is permitivity of free space.
Let, 'm' be the mass of an electron moving with velocity 'v' then centripetal force is given by,
Fc= mv2/r <------- (ii)
The electron revolve around the nucleus,
Fe = Fc
mv2/r = (1/4πε₀)(ze2/r2)
mv2 = (1/4πε₀)(ze2/r) <------ (iii)
From bohr's quantization condition, we have
mvr= nh/2π
v= nh/2πmr <------- (iv)
putting the value of v in equation (iii)
m(nh/2πmr)2= (1/4πε₀)(ze2/r)
[(nh)2/πmr]= (ze2/ε₀)
r= [ε₀. (nh)2]/(ze2πm)

For hydrogen atom
z=1
So,
r= (ε₀/πm) (nh/e)2 <------- (vi)
This is the required equation for radius of nth stationery orbit for H atom.

Velocity of nth stationery orbit for Hydrogen atom.


Bohr's assumed that hydrogen atom consists of unipositive charge concentrated on a nucleus and unit negative charge.

The electrons which revolve around the nucleus is balanced by centripetal force as shown in above figure.

Fc= mvn2/rn <------- (i)
The electrostatic force of attraction between electron and nucleus is given by,
Fe= (1/4πε₀)(ze2/r2) <------- (ii)
If vn be the velocity of electron in nth stationery orbit, the condition for electron to revolve in the circular orbit, the electrostatic force of attraction between electron and nucleus is provided by necessary centripetal force.
Fe= Fc
(1/4πε₀)(ze2/r2) = mvn2/rn
mvn2 = (1/4πε₀)(ze2/rn) <------ (iii)
From bohr's quantization condition, the angular momentum of an electron is given by,
mvnrn= nh/2π <------- (iv)
dividing equation (iii) by equation (iv) we get,
vn/rn=[(ze22π)/(4πε₀rnnh)]
vn= (ze22π/4πε₀rnnh) <------ (v)

For hydrogen atom , z=1
so,
vn= e2/2nhε₀ <------ (vi)
which is required expression of velocity of an electron in nth stationery orbit for hydrogen atom.
vn= (e2/2nε₀).(1/n)
vn= k. 1/n, where k=(e2/2nε₀)
vn ∝ 1/n
Hence, velocity of electron in an stationery orbit is inversely proportional to principle quantum number and decreases with ratio 1:2:3.

NOTE:
vn= (C/137).1/n
when n=1
v1= C/137
which is velocity of electron in ground state.

Total Energy of an electron in nth stationery orbit for hydrogen atom:

The total energy of an electron in a given orbit is the sum of Kinetic energy K.E. and Potential energy P.E. Let, Ek and Ep be the kinetic and potential energy of an electron from total energy of an electron is given by,

En= Ek + Ep <------- (i)
Let vn be the velocity of an electron in an stationery orbit then K.E. of an electron is given by,
Ek= (1/2)mvn2 <-------(ii)
where 'm' is mass of electron but,
mvn2= (1/4πε₀)(ze2/rn) <------ (iii)
Ek = 1/2(1/4πε₀)(ze2/rn) <------- (iv)
The potential energy of an electron is the amount of work done to bring an electron from infinity to rn i.e.
Ep= rn Fe. dr
Ep= rn (1/4πε₀)(ze2/r2).dr
Ep=(ze2/4πε₀) rn r-2.dr
Ep= -(1/4πε₀)(ze2/rn>) <---------- (v)
By using equation (iv) and equation (v) in equation (i) we get,
En= 1/2(1/4πε₀)(ze2/rn) -(1/4πε₀)(ze2/rn>)
(1/4πε₀)(ze2/rn) [(1/2)-1]
i.e. En= ze2/8πε₀rn <------ (vi)
But rn= (n2h2ε₀)/mπze2
So, equation (vi) becomes,
En= -ze2/(8πε₀)x[(n2h2ε₀)/mπze2]

For Hydrogen atom
z=1 so,
En= (-me4)/(8ε₀2n2h2)

It is the required expression for total energy of an electron in nth stationery orbit for hydrogen atom where '-ve' sign indicates that electron are bonded by nucleus.
En= -[(me4/8ε₀2h2)].1/n2
where, [(me4/8ε₀2h2)] is constant for all.




In case of any problem ask me in qustions section!!!!

© 2020 onlinelearnal.com