# How do I find mass equivalence of a photon?

The problem has a flame emitting a wavelength of $589\ \mathrm{nm}$, and asks for the mass of one photon of that wavelength. It tells me that $1\ \mathrm J = 1\ \frac{\mathrm{kg\cdot m^2}}{\mathrm s^2}$.

I converted to $\mathrm J$ using $E=\frac{h\cdot c}{\lambda}$. I'm not quite sure what to do past that. I have $3.36\cdot 10^{-19}\mathrm J$. I know the answer is supposed to be $3.75\cdot 10^{-34}$.

• Then the answer should be 3.75x10^-36 not to the -34. If that is what the book answer was then I don't know how they got that. – Chris Brown Jul 22 '19 at 0:32

$$E = \frac{h\cdot c}{\lambda}$$

$$1\ \mathrm J = 6.24150934 \times 10^{18}\ \mathrm{eV}$$

then you have $$E = mc^2 = \frac{h\cdot c}{\lambda}$$

Simply calculate the energy: $E=h\nu$

Then solve for momentum: $p=E/c$

Then divide by the speed of light: $p=mv$ or $m=p/v$

With the speed of light for $v$ so: $m=p/c$