An alternating source drives a series RLC circuit with an emf amplitude of 6.34 V, at a phase angle of +32.6°. When the potential difference across the capacitor reaches its maximum positive value of +5.08 V, what is the potential difference across the inductor (sign included)?

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m

its a circle not a square that's simple silly.

explanation: