Question

Determine the energy of the electron in hydrogen atom while revolving in nth orbit

Answer 1

The state of the atom wherein the electron is revolving in the orbit of smallest Bohr radius (a0) is the ‘Ground State’. In this state, the atom has the lowest energy. The energy in this state is:

E1 = -13.6 eV

Answer 2

Answer:

When electron is revolving around the electron, it has kinetic energy. Since, it is attracted towards the nucleus by electrostatic force of attraction, it has potential energy. Therefore, the sum of kinetic energy and potential energy gives the total energy.

We know, Kinetic energy of electron in nth orbit is given by:

KE=12mvn2……..(i)

KE=12mvn2……..(i)

We know from previous topic,

vn=e22ε∘nh

vn=e22ε∘nh

So, putting the value of Vn in equation (i), we get:

KE=12m(e22ε∘nh)2

KE=12m(e22ε∘nh)2

∴KE=me48ε∘2n2h2

∴KE=me48ε∘2n2h2

Similarly, Potential energy (P.E.) = Potential due to nucleus at distance r × charge of electron i.e.,

PE=e4πε∘rn×(−e)

PE=e4πε∘rn×(−e)

or,PE=−14πε∘e2rn

or,PE=−14πε∘e2rn

We know from the previous topic,

rn=ε∘n2h2πme2

rn=ε∘n2h2πme2

Then Potential energy becomes,

PE=−e24ε∘πme2ε∘n2h2

PE=−e24ε∘πme2ε∘n2h2

∴PE=−me44ε∘2n2h2

∴PE=−me44ε∘2n2h2

Now, Total Energy(En)=Kinetic Energy + Potential Energy

Now, Total Energy(En)=Kinetic Energy + Potential Energy

or,En=me48ε∘2n2h2–me44ε∘2n2h2

or,En=me48ε∘2n2h2–me44ε∘2n2h2

∴En=me48ε∘2n2h2

∴En=me48ε∘2n2h2

Putting the values of m, e, and h in above equation as;

m = 9.1×10-31 ­­kg

e = 1.6×10-19 C

εo = 8.85×10-12 F/m

h = 6.62×10-34Js we get,

∴ Total Energy of nth orbit = – 13.6/n2 eV…….(iii)