23P1Q28

28 A

$\displaystyle \begin{aligned} {{E}_{photon}}&={{E}_{H}}-{{E}_{L}} \\ \frac{(6.63\times {{10}^{-34}})(3.00\times {{10}^{8}})}{\lambda }&=(122.4-30.6)(1.60\times {{10}^{-19}}) \\ \lambda &=1.35\times {{10}^{-8}}\text{ m} \end{aligned}$

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