Answer to Question #184424 in Organic Chemistry for Rupesh

Question

derive the expression for the total energy of an electron in the nth orbit

Answer 1

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

$KE=\frac{1}{2} mv_n2...........(1)$

Also, we know that,

$vn=\frac{e^2}{2nh}ε°$

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

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

$K.E = \frac{me^4}{8n^2h^2} (εο)^2$

In the same way,

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

We know,

$rn = ε^∘n^2h^2πme^2$

Now, Potential energy =

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

Hence, $PE = −\frac{me^4}{4}ε^∘2n^2h^2$

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,

$m= 9.1 ×10^{-31}$

$e = 1.6 × 10^{-19}$

$ε^o = 8.85 ×10^{-12}$

$h = 6.62 × 10^{-34}$

  we get,

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