# Question #19605

May 12, 2016

$\Delta E = - {R}_{H} \cdot {\left(Z\right)}^{2} \left[\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right]$

#### Explanation:

In the hydrogen atom, the energy of the electron in a given energy level is given by : ${E}_{n} = - {R}_{H} \cdot {\left(\frac{Z}{n}\right)}^{2}$

${E}_{f} = - {R}_{H} \cdot {\left(\frac{Z}{n} _ f\right)}^{2}$

${E}_{i} = - {R}_{H} \cdot {\left(\frac{Z}{n} _ i\right)}^{2}$

$\Delta E = {E}_{f} - {E}_{i}$

$\Delta E = \left[- {R}_{H} \cdot {\left(\frac{Z}{n} _ f\right)}^{2}\right] - \left[- {R}_{H} \cdot {\left(\frac{Z}{n} _ i\right)}^{2}\right]$

take $- {R}_{H} \cdot {\left(Z\right)}^{2}$ as a common factor,

$\Delta E = - {R}_{H} \cdot {\left(Z\right)}^{2} \left[\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right]$

The energy of the photon emitted is given by:

$\Delta E = - h \frac{c}{\lambda}$

please note that a negative sign must be introduced to the energy expression since energy is released.

combining the two equations gives:

$- h \frac{c}{\lambda} = - {R}_{H} \cdot {\left(Z\right)}^{2} \left[\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right]$ (equation 1)

$h \text{ is Planck's constant} = 6.626 \cdot {10}^{-} 34 J . s$
${R}_{H} \text{ is Rydberg constant} = 2.178 \cdot {10}^{-} 18 J$
$Z \text{ is the atomic number of the hydrogen atom} = 1$
$n \text{ is principle quantum number}$

${n}_{i} \text{ is the initial quantum state of the electron.}$
${n}_{f} = 2$

plugging the numbers in (equation 1)

$- \frac{6.626 \cdot {10}^{-} 34 J . s \cdot 2.998 \cdot {10}^{8} \frac{m}{s}}{397.2 \cdot {10}^{-} 9 m} = - 2.178 \cdot {10}^{-} 18 J \cdot {\left(1\right)}^{2} \left[\frac{1}{2} ^ 2 - \frac{1}{n} _ {i}^{2}\right]$

$- \frac{6.626 \cdot {10}^{-} 34 \cancel{J} . \cancel{s} \cdot 2.998 \cdot {10}^{8} \frac{\cancel{m}}{\cancel{s}}}{397.2 \cdot {10}^{-} 9 \cancel{m}} = - 2.178 \cdot {10}^{-} 18 \cancel{J} \cdot {\left(1\right)}^{2} \left[\frac{1}{2} ^ 2 - \frac{1}{n} _ {i}^{2}\right]$

solve for ${n}_{i}$,

${n}_{i} = 7$