# What is the difference in energy between

What is the difference in energy between the two levels responsible for the ultraviolet emission line of the magnesium atom at 285.2 nm?

By commons: (User:Pumbaa User:Greg Robson) [CC BY-SA 2.0 uk], via Wikimedia Commons

#### Solution:

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We will first figure out the frequency, and use that value to calculate the difference in energy.

The first equation we will use is:

$v\,=\,\dfrac{c}{\lambda}$

(Where $v$ is frequency, $c$ is speed of light, and $\lambda$ is wavelength)

The question states that the line of emission for the magnesium atom is at 285.2 nm. We will write this value in meters.

$285.2$ nm = $2.852 \times 10^{-7}$ m

Let us substitute the values into our equation to figure out the frequency.

$v\,=\,\dfrac{2.998 \times 10^8 \text{m/s}}{2.852 \times 10^{-7}\text{m}}$

(The speed of light is equal to $2.998 \times 10^8$ m/s)

$v\,=\,1.05 \times 10^{15}$ /s

We will now use the following formula to figure out the energy.

$E\,=\,hv$

(Where $E$ is energy, $h$ is the Planck’s constant, and $v$ is the frequency)

$E\,=\,(6.626\times 10^{-34}\text{J}\cdot \text{s})(1.05 \times 10^{15}\text{/s})$

$E\,=\,6.96\times 10^{-19}$ J

The difference in energy is $6.96\times 10^{-19}$ J