Chapter 28.1, Problem 7PP

To determine

To Calculate: Energy of the excited state.

Answer to Problem 7PP

  $\text{ -50}\text{.32 eV}$

Explanation of Solution

Given:

Ground state energy of helium ion, $E_g=-54.4\text{eV}$

Wavelength of the emitted photon, $\lambda =304\text{nm}$

Formula Used:

Transition energy, $\Delta E=E-E_g$

  $\Delta E=\frac{hc}{\lambda }$

Here, h is the Planck’s constant, c is the speed of the light and $\lambda$ is the wavelength of the emitted of photon.

Calculations:

Substitute the values and solve for E .

  $\begin{array}{l}E=E_g+\Delta E\\ E=-54.4\text{eV}\times \text{1}\text{.602}\times {\text{10}}^{-19}\frac{\text{J}}{\text{eV}}\text{+}\frac{6.62\times {10}^{-34}\text{J}\cdot \text{s}\times 3\times {10}^8\text{m/s}}{304\times {10}^{-9}\text{m}}\\ E=-87.15\times {10}^{-19}\text{J}+6.53\times {10}^{-19}\text{J}\\ E=-\text{80}\text{.62}\times {10}^{-19}\text{J = -50}\text{.32 eV}\end{array}$

Conclusion:

Thus, the energy of the excited state is $\text{ -50}\text{.32 eV}$ .