Question

A hydrogen atom has orbital angular momentum $3.65\times {10}^{-34}Js$

a. What letter $\left(s,p,dorf\right)$ describes the electron?

b. What is the atom’s minimum possible energy? Explain.

Short answer

a. The electron is $L=3$.

b. The atom’s minimum possible energy is$-0.850eV$

Step-by-step solution

Part (a) step 1: Given Information 

We need to describe the electron letter is $\left(s,p,dorf\right)$.

We have given angular momentum is$3.65\times {10}^{-34}Js$.

Part (a) step 2: Simplify

Finding orbital angular momentum is:

$L=\sqrt{l\left(l+1\right)}ħ$

$\frac{L}{ħ}=\frac{\sqrt{l\left(l+1\right)}ħ}{ħ}$ (Divide both sides by $h$) $\begin{array}{rr}\frac{L}{ħ}& =\sqrt{l\left(l+1\right)}\\ {\left(\frac{L}{ħ}\right)}^2& ={\left(\sqrt{l\left(l+1\right)}\right)}^2\\ {\left(\frac{L}{ħ}\right)}^2& =l\left(l+1\right)\end{array}$

$$\begin{gathered} {\left(\frac{\left(3.65\times {10}^{-34}Js\right)}{\left(1.5\times {10}^{-34}Js\right)}\right)}^2=l\left(l+1\right) \\ \begin{array}{c}12.08=l\left(l+1\right)\\ \Rightarrow l=3\end{array} \end{gathered}$$

Part (b) step 1: Given Information 

We need to find the atom’s minimum possible energy.

Part (a) step 2: Simplify

Finding the atom minimum energy is :

$$\begin{gathered} E_{min}=-\frac{13.60eV}{n^2}\left(wheren=4\right) \\ {E}_{min}=-\frac{13.60eV}{4^2} \\ {E}_{min}=-\frac{13.60eV}{16} \\ {E}_{min}=-0.850eV \end{gathered}$$