# Chapter 6 - Electronic Structure of Atoms - Additional Exercises: 6.89b

The wavelength of light that must be absorbed to achieve that transition is $\lambda\approx91.12nm$

#### Work Step by Step

1) Calculate the change in energy for the transition from $n=1$ to $n=\infty$ $\Delta E= E_f-E_i$ $\Delta E=(-2.18\times10^{-18}J)\Big(\frac{1}{n_f^2}-\frac{1}{n_i^2}\Big)$ $\Delta E=(-2.18\times10^{-18}J)\Big(\frac{1}{\infty^2}-\frac{1}{1^2}\Big)$ $\Delta E=(-2.18\times10^{-18}J)(0-1)$ $\Delta E=2.18\times10^{-18}J$ 2) We know that the energy of a photon must equal the change in energy for the transition, since to move an atom from $n=1$ to $n=\infty$, a photon with an energy must be absorbed. In other words, $$E_p=|\Delta E|$$Therefore, $$E_p=2.18\times10^{-18}J$$ 3) Find the wavelength of light that must be absorbed The wavelength of light that must be absorbed $(\lambda)$ is calculated according to the formula $$E_p=h\frac{c}{\lambda}$$$$\lambda=\frac{h\times c}{E_p}$$ We have Planck's constant $h\approx6.626\times10^{-34}J.s$ and the speed of light in vacuum $c\approx2.998\times10^8m/s$ Therefore, $$\lambda=\frac{(6.626\times10^{-34}J.s)\times(2.998\times10^8m/s)}{2.18\times10^{-18}J}$$$$\lambda\approx9.112\times10^{-8}m\approx91.12nm$$

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