Special Relativity: Energy-momentum Light Relation

Classical Mechanics Level 2

Einstein's equation $E=mc^2$ is well known but is not the full story. The full equation is $E^2 = p^2c^2 + m^2c^4$ . We acquire Einstein's formula for systems at rest; however, we obtain the formula $E=pc$ for systems without mass.

Determine the momentum carried by a blue photon with wavelength ( $\lambda$ ) of $450 \text{ nm}$ and $h = 6.626 * 10^{-34} \text{ Js}$ , in units of $\text{Newton-seconds}$ using the following formulas: $E = \frac{hc}{\lambda} \\ E = pc$

David's Special Relativity Set

1.47 * 10^{-27} \text{Ns}

1.48 * 10^{-36} \text{Ns}

1.47 * 10^{-36} \text{Ns}

1.48 * 10^{-27} \text{Ns}

1 solution

David Hontz
Dec 22, 2016

$E=pc \text{ and } E =\frac{hc}{\lambda} \rightarrow pc = \frac{hc}{\lambda} \rightarrow p=\frac{h}{\lambda} \\ p = \frac{6.626 * 10^{-34} \text{ Js}}{450 * 10^{-9} \text{ m}} = 1.47 * 10^{-27} \frac{\text{Nms}}{\text{m}} = \boxed{1.47 * 10^{-27} \text{ Ns}}$

Special Relativity: Energy-momentum Light Relation

David's Special Relativity Set

1 solution

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