## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

The energy of the light is given by $$\Delta = \frac{hc}{\lambda}$$ The energy of a particle in the box is given by $$E_n = {n^2}\frac{h^2}{8mL^2}$$ Now the minimum energy required is $$E_n = E_2 - E_1 = \frac{4h^2}{8mL^2} - \frac{h^2}{8mL^2} = \frac{3h^2}{8mL^2}$$ The, equating the energy difference yields $$L^2=\frac{3h\lambda}{8mc}=\frac{3 \times1.38\times10^{-23}\times600\times10^{-9}}{8\times9.1\times10^{31}\times3\times10^8}=1.73\times10^{-8} \text m$$