## Related questions with answers

A hydrogen atom is in its first excited state (n = 2). Using the Bohr theory of the atom, calculate the potential energy.

Solutions

VerifiedIn this problem, a hydrogen atom is excited to the first excited state, corresponding to $n = 2$. We use the Bohr model in this problem. For this part, we calculate the electrical potential energy of the atom.

Given values:

$n=2$ $r=0.212 \ nm=0.212 \cdot 10^{-9} \ m$

In part $\textbf{e)}$ we have to find the potential energy.

From the problem $\textit{11(a)}$, for the state $n=2$, we know that $r_{n=2}=0.212\textrm{nm}=0.212\times10^{-9}\textrm{m}$. Thus solving for potential energy $P.E$, we get:

$\begin{align*} P.E_{n=2}=-\left(1.60\times10^{-19}\right)^{2}\frac{k}{r_{n=2}} \end{align*}$

It follows:

$\begin{align*} P.E_{n=2}=-\left(1.60\times10^{-19}\right)^{2}\frac{8.99\times10^{9}\frac{\textrm{N}\cdot{\textrm{m}^{2}}}{\textrm{C}^{2}}}{0.212\times10^{-9}\textrm{m}}=-1.09\times10^{-18}\textrm{J} \end{align*}$

Solution is:

$\begin{align*} \boxed{P.E_{n=2}=-6.8\textrm{eV}} \end{align*}$

$PE_2 = \dfrac{k \ (-e) \ (e)}{r_2}$

$= \dfrac{(8.99e9) \ (-1.6e-19) \ (-1.6e-19)}{0.212e-9}$

$= -1.09e-18 \ J$

$= -6.80 \ eV$

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