What is the energy of a photon emitted with a wavelength of 448 nm?

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What is the energy of a photon emitted with a wavelength of 448 nm?

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Answer 1 · 2016-06-15T02:58:01

$4.44 \cdot 10^{- 19} J$ $4.44 * 10^-19J$

Explanation:

A photon is a particle of light that is released when an electron transitions from a high energy state to a low energy state.

The energy associated with the process is given by the following equation:

$E = h \cdot f$ $E=h*f$

where $E$ $E$ is the energy associated with the photon, $h$ $h$ is Planck's constant ( $6.63 \cdot 10^{- 34} J \cdot s)$ $6.63*10^-34J*s)$ , and $f$ $f$ is the characteristic frequency associated with the photon (expressed in Hertz, or $s^{- 1}$ $s^-1$ ).

With the information given in the problem, it might at first seem as if the problem cannot be solved. However, frequency, $f$ $f$ , can be related to wavelength, $λ$ $lambda$ , in the following manner:

$f = \frac{c}{λ}$ $f=c/lambda$

Since the speed of light, $c$ $c$ , is relatively constant ( $3.00 \cdot 10^{8} \frac{m}{s}$ $3.00*10^8m/s$ ) (this is true for a vacuum, but light slows down in other mediums) and wavelength is known, the second equation can be substituted into the first:

$E = \frac{h \cdot c}{λ}$ $E=(h*c)/lambda$
$E = \frac{(6.63 \cdot 10^{- 34} J \cdot s) \cdot (3.00 \cdot 10^{8} \cdot \frac{m}{s})}{448 \cdot 10^{- 9} \cdot m}$ $E=((6.63*10^-34J*s)*(3.00*10^8*m/s))/(448*10^-9*m)$
$E = 4.44 \cdot 10^{- 19} J$ $E=4.44*10^-19J$

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What is the energy of a photon emitted with a wavelength of 448 nm?

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