Question

Derive the expression for the total energy of an electron in the nth Orbit​

Answer 1

Answer:

This may be the answer for u r question

Answer 2

Total Energy of nth orbit = – 13.6/n2 eV

Explanation:

Total energy of an electron in nth orbit of an atom is sum of the Kinetic Energy and Potential energy in the orbit.

∴ $E_n = KE_n + PE_n$

We know,

Kinetic energy of electron in nth orbit can be given as -

$KE=\frac{1}{2} mv_n2$……..(i)

Also, we know that,

$vn=\frac{e^2}{2nh}$ε°

∴ By putting the value of Vn in equation (i), we get -

 $KE = \frac{1}{2}m(\frac{ e^2}{2nh})^2$εο

∴$K.E = \frac{me^4}{8n^2h^2}$(εο$)^2$

In the same way,

Potential energy (P.E.) = −14πε∘e × 2rn

We know,

rn = ε∘$n^2h^2$π$me^2$

Now, Potential energy =

  PE = $-\frac{e^2}{4}$ε∘π$me^2$ε∘$n^2h^2$ ..........(ii)

∴ PE = −$\frac{me^4}{4}$ε∘2$n^2h^2$

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

En = $\frac{me^4}{8}$ε∘$2n^2h^2$–$\frac{me^4}{4}$ε∘$2n^2h^2$

∴ En =$\frac{me^4}{8}$ε∘$2n^2h^2$

Putting the values of m, e, and h in above equation we get,

m = 9.1 ×$10^{-31}$ ­­kg

e = 1.6 × $10^{-19}$ C

εo = 8.85 ×$10^{-12}$ F/m

h = 6.62 × $10^{-34}$Js we get,

∴ Total Energy of nth orbit = – 13.6/n2 eV