# What is the energy of an emitted flame with a wavelength of 6.5 * 10^-20 m?

Feb 2, 2016

$3.058 \times {10}^{- 6} j$

#### Explanation:

we know that,
$E = h f$

$\mathmr{and} , E = h \frac{c}{\lambda}$[as, $c = f \lambda$, so, $f = \frac{c}{\lambda}$]

$= 6.626 \times {10}^{- 34} \cdot \frac{3 \cdot {10}^{8}}{6.5 \cdot {10}^{-} 20}$[$h$=plunk's constant, $c$= speed of light]

$= 3.058 \times {10}^{- 6} j$

Feb 2, 2016

The wavelength is incredibly short!!! It is difficult to imagine how did they measure it!

#### Explanation:

Here you have a flame of quite short wavelength well beyond the gamma ray range!!!
If you want, I think, you could use Einstein's equation for the energy $E$ of a photon of a certain frequency $\nu$:
$E = h \cdot \nu$
where:
$k = 6.63 \times {10}^{-} 34 J s$ is Planck's constant;
and also frequecy is related to eavelength $\lambda$ through the speed of light $c$ as:
$c = \lambda \cdot \nu$
giving an energy of:
$E = 6.63 \times {10}^{-} 34 \cdot \frac{3 \times {10}^{8}}{6.5 \times {10}^{-} 20} = 3.06 \times {10}^{-} 6 J$ and this is for a photon only!!!