Question

Consider an atom with two closely spaced excited states A and B. If the atom jumps to ground state from A or from B, it emits a wavelength of 500 nm or 510 nm, respectively. What is the energy difference between states A and B?

Short answer

The energy difference between states A and B is 0.049 eV .

Step-by-step solution

The given data

1. Wavelength of the emitted atom when it jumps from A,${\mathrm{\lambda }}_{\mathrm{A}}=500\mathrm{nm}$
2. Wavelength of the emitted atom when it jumps from B,${\mathrm{\lambda }}_{\mathrm{B}}=510\mathrm{nm}$

Understanding the concept of plank’s relation:

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

From Planck's relation, get the individual energies of each exciting case that is the excitation of the atom either from A or B. Now, the difference value of these individual energies is the required energy difference.

Formula:

The energy of the photon due to Planck’s relation,

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (1)

Here, consider the known data below.

The Plank’s constant is,

$$\begin{gathered} \mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ =\left(6.242\times {10}^{18}\times 6.63\times {10}^{-34}\right)\mathrm{eV}.\mathrm{s} \\ =41.384\times {10}^{-16}\mathrm{eV}.\mathrm{s} \end{gathered}$$

The speed of light is,

$$\begin{gathered} \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \\ =\left(3\times {10}^8\times {10}^9\right)\mathrm{nm}/\mathrm{s} \\ =3\times {10}^{17}\mathrm{nm}/\mathrm{s} \end{gathered}$$

Calculation of the energy difference:

Using the given data and equation (1), the energy difference between the two given states A and B can be calculated as follows:

$$\begin{gathered} ∆\mathrm{E}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{A}}}-\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{B}}} \\ =\mathrm{hc}\left[\frac{1}{{\mathrm{\lambda }}_{\mathrm{A}}}-\frac{1}{{\mathrm{\lambda }}_{\mathrm{B}}}\right] \\ ∆\mathrm{E}=41.384\times {10}^{-16}\mathrm{eV}.\mathrm{s}\times 3\times {10}^{17}\mathrm{nm}/\mathrm{s}\left[\frac{1}{500\mathrm{nm}}-\frac{1}{510\mathrm{nm}}\right] \\ =\left(1240\mathrm{eV}.\mathrm{nm}\right)\left[\frac{0.002}{\mathrm{nm}}-\frac{0.0019}{\mathrm{nm}}\right] \\ =\left(1240\mathrm{eV}.\mathrm{nm}\right)\left(\frac{4\times {10}^{-5}}{\mathrm{nm}}\right) \\ =0.049\mathrm{eV} \end{gathered}$$

Hence, the value of the energy difference is 0.049 eV .