Two airplanes leave an airport at the same time. The velocity of the first airplane is
680 m/h at a heading of 57.7◦. The velocity
of the second is 580 m/h at a heading of 119◦. How far apart are they after 2.4 h? Answer in units of m.

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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