The transition of an electron from a higher level to a lower level results in the emission of a photon of wavelength $350.0\ \mathrm{nm}$. If the energy of the higher level is $-3.24\times10^{-19}\ \mathrm J$, calculate the energy of the lower level.

By using $E = h\nu$

I find $E = 5.679\times10^{-28}\ \mathrm J$

How do I use this value to solve the qus?


All you need are two equations and two constants (speed of light and Planck constant): \begin{eqnarray} \mathrm{E} & = & h\nu\\ h & = & \mathrm{6.62606957 \cdot 10^{−34} J\cdot s}\\ \mathrm{c} & = & \lambda\nu\\ \mathrm{c} & = & 299792458\ \mathrm{m\cdot s}^{-1}\\ \end{eqnarray}

Using these data, the energy of a photon with $\lambda = 350\ \mathrm{nm}$ calculates to:

$$\mathrm{E} = \frac{6.62606957 \cdot 10^{-34} \cdot 2.99792458 \cdot 10^8}{350 \cdot 10^{-9}} \cdot \frac{\mathrm{J \cdot s \cdot m}}{\mathrm{s \cdot m}} = 0.05676 \cdot 10^{-17} \mathrm{J} = 5.676 \cdot 10^{-19} \mathrm{J}$$

This is the energy difference to the higher level.

  • $\begingroup$ Thanks for your reply @Klaus Warzecha. Am I right to say that after substituting wavelength = 350nm into the equations, I'm already finding the energy of the lower level? $\endgroup$
    – Kenny
    Nov 17 '14 at 14:41
  • $\begingroup$ @Kenny Almost, you're calculating the energy of a photon with $\lambda$ = 350 nm. This is the difference to the other level for which the energy was given. $\endgroup$ Nov 17 '14 at 18:04
  • $\begingroup$ Thank you for your reply. I got totally messed up with my units. $\endgroup$
    – Kenny
    Nov 18 '14 at 4:43
  • $\begingroup$ @Kenny My pleasure :) Units are a pretty helpful thing - unless one confuses imperial and metric measures and therefore looses a Mars Climate orbiter. Fortunately, NASA openly stands to this historical error and even cites it educational materials. Good on them! $\endgroup$ Nov 18 '14 at 9:04

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