Question

(i) The energy associated with the first orbit in the hydrogen atom is $2.18\times {10}^{-18}J/atom$.
What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bhor’s fifth orbit for the hydrogen atom.

Answer 1

Hint: The formula to calculate the energy associate with nth orbit of the hydrogen atom is
$$E_n={\displaystyle \frac{-2.18\times {10}^{-18}}{n^2}}J/atom$$
Where $E_n$ = energy associated with nth orbit.
The formula to calculate the radius of the Bhor’s nth orbit is
$r_n=0.0529n^2$
Where $r_n$ = Radius of the nth orbit
n = orbit number or Principal quantum number.

Complete step by step answer:
- From the question we have to calculate the energy associated with 5th orbit and radius of Bhor’s 5th orbit.
(i) Energy associated with 5th orbit of hydrogen atom is ( n = 5)
$$\begin{array}{rl}& E_n={\displaystyle \frac{-2.18\times {10}^{-18}}{n^2}}J/atom\\ & ={\displaystyle \frac{-2.18\times {10}^{-18}}{5^2}}\\ & =-8.68\times {10}^{-20}J/atom\end{array}$$
- Therefore the energy associated with the 5th orbit of the hydrogen atom is $-8.68\times {10}^{-20}$ J/atom .

(ii) The radius of the Bhor’s 5th orbit of hydrogen atom is ( n = 5)
$$\begin{array}{rl}& r_n=0.0529n^2\\ & r_5=0.0529\times {(5)}^2\\ & r_5=1.3225nm\end{array}$$
- Therefore the radius of the 5th Bhor’s orbit of the hydrogen atom is 1.3225 nm or 13.225$\stackrel{o}{\text{A}}$.

Note: As the principal quantum number (n) values increases the radius of the orbit increases because they are directly proportional to each other. As the principal quantum number increases the energy of the orbital decreases because they are inversely proportional to each other. The energy is going to express in the form of Joule/atom and the radius is going to express in the form of nm (nanometer) or Angstrom.
1 nm = 10 $\stackrel{o}{\text{A}}$ .