Energy

source : abyss.uoregon.edu

## Energy and velocity of Electron-

## Calculation for the energy of electron in each orbit –

*The energy of a moving electron is the sum of kinetic and potential energy. The total energy of the electron is ‘E’.*

*KE = 1/2 mv ^{2}*

*Potential energy = – KZe ^{2}/r*

*E = Kinetic energy + Potential energy*

*= 1/2 mv ^{2} – KZe^{2}/r ——– eq 1*

*We know,*

*KZe ^{2} /r = mv^{2} ———— eq 2*

*Putting the value of mv2 from eq (2) to eq (1)*

*E = KZe ^{2} /2r – KZe^{2}/r = -KZe^{2}/2r*

*We know,*

*r = n ^{2}h^{2}/4π^{2}me^{2}KZ*

*so,*

*E = (-KZe ^{2} x 4π^{2}me^{2}KZ /2 n^{2} h^{2}) = – 2 K^{2}Z^{2} π^{2}me^{4} /n^{2} h^{2}*

*For H-atom, Z=1*

*therefore,*

*E = – 2 K*^{2} **π**^{2}me^{4} /n^{2} h^{2}

^{2}

**π**

^{2}me^{4}/n^{2}h^{2}*h = 6.625 x 10 ^{-34} , π = 3.14 , m = 9.1 x 10^{-31}, e = 1.6 x 10^{-19} , K = 9 x 10^{9} Nm^{2}/C^{2}*

*putting the values of π , m , e, K , and h,*

*E = – 2 x 9 x 10^{9} x 9 x 10^{9} x 3.14 x 3.14 x 9.1 x 10^{-31} x (1.6 x 10^{-19} )^{4} / n^{2} x 6.625 x 10^{-34} x 6.625 x 10^{-34}*

*E = -21.70 x 10*^{-19} / n^{2} joule

^{-19}/ n

^{2}joule

*For ,n=1 , E = – 21.70 x 10*^{-19} / 1 x 1= – 21.70 x 10^{-19} J = – 21.70 x 10^{-12} erg = -13.6 eV

^{-19}/ 1 x 1= – 21.70 x 10

^{-19}J = – 21.70 x 10

^{-12}erg = -13.6 eV

*If, n=2 , E = – 21.70 x 10*^{-19} / 2 x 2= – 5.42 x 10^{-19} J

^{-19}/ 2 x 2= – 5.42 x 10

^{-19}J

*For n=3 , E = – 21.70 x 10*^{-19} / 3 x 3= – 2.41 x 10^{-19} J

^{-19}/ 3 x 3= – 2.41 x 10

^{-19}J

## Energy of an electron in lower orbit (n_{1}) and higher orbit(n_{2})-

*E _{n1} = – 2 K^{2}Z^{2} π^{2}me^{4} /n_{1}^{2} h^{2}*

*For H – atom,*

*Z =1*

*E _{n1} = – 2 K^{2} π^{2}me^{4} /n_{1}^{2} h^{2}*

*E _{n2} = – 2 K^{2} π^{2}me^{4} /n_{2}^{2} h^{2}*

*E _{n2} – E_{n1} = – 2 K^{2} π^{2}me^{4} /n_{2}^{2} h^{2} + 2 K^{2} π^{2}me^{4} /n_{1}^{2} h^{2}*

*Δ E = (2 K ^{2} π^{2}me^{4} / h^{2 })[( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]*

*According to Planck’s relation,*

*Δ E = hc/ λ *

*hc/ λ = (2 K^{2} π^{2}me^{4} / h^{2})[( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]*

*1 / λ =( 2 K2 π^{2}me^{4} / ch^{3})[( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]*

*K = 1 in CGS unit*

**[2 π ^{2}me^{4} / ch^{3}] was found to be 1.096 x 10^{5} cm^{-1} which is equal to Rydberg constant “R_{H}“**

*1 / ***λ = R**_{H} [( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]

**λ = R**

_{H}[( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]*1 / ***λ = 1.096 x 10**^{5} [( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]

**λ = 1.096 x 10**

^{5}[( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]## Calculation of velocity of electron-

*According to one postulate of Bohr theory ( Electron can revolve in those orbits whose angular momentum is ‘h / 2 π’ or its integral multiple).*

*mvr = nh/2 π *

*v = nh/2 πmr ——–eq 1*

*We Know,*

*r = n ^{2}h^{2}/4π^{2}me^{2}KZ*

*For hydrogen , Z = 1 ,hence,*

*r = n ^{2}h^{2}/4π^{2}me^{2}K ———- eq2*

*putting the value of ‘r’ from eq (2) to eq (1)*

*v = nh x 4 π^{2}me^{2}K / 2πmn^{2}h^{2}*

*v = 2 π e^{2}K / nh*

*h = 6.625 x 10 ^{-34} , π = 3.14 , e = 1.6 x 10^{-19} , K = 9 x 10^{9} N m^{2}/C^{2}*

*v = 2 x 3.14 x 1.6 x 10^{-19} x 1.6 x 10^{-19} x 9 x 10^{9} /n x 6.625 x 10^{-34}*

*v = 21.84 x 10*^{5}/ n metre = 21.84 x 10^{7} /n cm = 2.184 x 10^{8} / n cm.

^{5}/ n metre = 21.84 x 10

^{7}/n cm = 2.184 x 10

^{8}/ n cm.