Photon energy:

$\overline{){\mathbf{E}}{\mathbf{=}}{\mathbf{h}}{\mathbf{f}}{\mathbf{=}}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda}}}$

1MHz = 10^{6}Hz

1eV = 1.6 × 10^{-19} J

f = 49.0MHz(10^{6}Hz/1MHz) = 49.0 × 10^{6} Hz

Calculate the energy, in electron volts, of a photon whose frequency is the following.

49.0 MHz

____ eV

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