Energy source : abyss.uoregon.edu

## 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 mv2

Potential energy = – KZe2/r

E =  Kinetic energy + Potential energy

= 1/2 mv2 – KZe2/r         ——– eq 1

We know,

KZe2 /r = mv2           ———— eq 2

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

E = KZe2 /2r – KZe2/r = -KZe2/2r

We know,

r = n2h2/4π2me2KZ

so,

E = (-KZe2 x  4π2me2KZ /2 n2 h2) = – 2 K2Z2 π2me4 /n2 h2

For H-atom, Z=1

therefore,

### E = – 2 K2 π2me4 /n2 h2

h = 6.625 x 10-34 , π = 3.14 , m = 9.1 x 10-31, e = 1.6 x 10-19 , K = 9 x 109 Nm2/C2

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

E = – 2 x 9 x 109 x 9 x 109 x 3.14 x 3.14 x 9.1 x 10-31 x (1.6 x 10-19 )4 / n2 x 6.625 x 10-34 x 6.625 x 10-34

## Energy of an electron in lower orbit (n1) and higher orbit(n2)-

En1 = – 2 K2Z2 π2me4 /n12 h2

For H – atom,

Z =1

En1 = – 2 K2 π2me4 /n12 h2

En2 = – 2 K2 π2me4 /n22 h2

En2 – En1 = – 2 K2 π2me4 /n22 h2 + 2 K2 π2me4 /n12 h2

Δ E = (2 K2 π2me4 / h2 )[( 1/ n12) – ( 1/ n22)]

According to Planck’s relation,

Δ E = hc/λ

hc/λ = (2 K2 π2me4 / h2)[( 1/ n12) – ( 1/ n22)]

1 / λ =( 2 K2 π2me4 / ch3)[( 1/ n12) – ( 1/ n22)]

K = 1 in CGS unit

[2 π2me4 / ch3] was found to be 1.096 x 105 cm-1 which is equal to Rydberg  constant “RH

## 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/2π

v = nh/2πmr        ——–eq 1

We Know,

r = n2h2/4π2me2KZ

For hydrogen , Z = 1 ,hence,

r = n2h2/4π2me2K        ———- eq2

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

v = nh x 4π2me2K / 2πmn2h2

v = 2π e2K / nh

h = 6.625 x 10-34 , π = 3.14 ,  e = 1.6 x 10-19 , K = 9 x 109 N m2/C2

v = 2 x 3.14 x 1.6 x 10-19 x 1.6 x 10-19 x 9 x 109 /n x 6.625 x 10-34