Question

Two excited energy levels are separated by the very small energy difference $∆E$. As atoms in these levels undergo quantum jumps to the ground state, the photons they emit have nearly identical wavelengths $\lambda$.

a. Show that the wavelengths differ by

$∆\lambda =\frac{{\lambda }^2}{hc}∆E$

b. In the Lyman series of hydrogen, what is the wavelength difference between photons emitted in the $n=20$to $n=1$transition and photons emitted in the $n=21$to $n=1$transition?

Short answer

The change in wavelength is $\frac{{\lambda }^2}{hc}∆E$and the wavelength difference of two transitions is $0.021nm$.

Step-by-step solution

Part (a) Step 1:

Use the relationship between the energy of the photon and wavelength.

The expression for the energy of the photon is,

$E=\frac{hc}{\lambda }$

Here, $h$is the planck's constant and $c$is the speed of light.

Differentiate the equation $E=\frac{hc}{\lambda }$with respect to $\lambda$.

$$\begin{gathered} \frac{dE}{d\lambda }=-\frac{hc}{{\lambda }^2} \\ d\lambda =-\frac{{\lambda }^2}{hc}dE \end{gathered}$$

Rewrite the above equation for small change in energy corresponding to small change in wavelength. So,

role="math" localid="1648727445072" $∆\lambda =\frac{{\lambda }^2}{hc}∆E$

Therefore, the changes in wavelength is$\frac{{\lambda }^2}{hc}∆E$.

Part (b) Step 1:

Use the result of part (a) to find the difference of wavelength between two transitions.

The energy difference between the states $21$and $20$is,

$$\begin{gathered} ∆E=E_{21}-E_{20} \\ =-\frac{13.6eV}{{21}^2}-\left(-\frac{13.6eV}{{20}^2}\right) \\ =\frac{13.6eV}{{20}^2}-\frac{13.6eV}{{21}^2} \\ =3.16\times {10}^{-3}eV \end{gathered}$$

The energy of emitting photon when transition between the states $20$and $1$is.

$$\begin{gathered} E_{20\to 1}=\left(-\frac{13.6eV}{{20}^2}\right)-\left(\frac{-13.6eV}{I^2}\right) \\ =13.56eV \end{gathered}$$

The wavelength of emitting photon when transition between the states $20$and $1$is.

$\lambda =\frac{hc}{E_{20\to 1}}$

Substitute $240eV.nm$for $hc$and $13.56eV$for .

$$\begin{gathered} \lambda =\frac{1240eV.nm}{13.56eV} \\ =91.4nm \end{gathered}$$

Here the wavelength of the emitted photon for transition $20\to 1$is almost equal to the wavelength of the emitted photon for transition$21\to 1$.

Difference of wavelength;

The difference between two wavelength is ,

$∆\lambda =\frac{{\lambda }^2}{hc}∆E$

Substitute $91.4nm$for $\lambda$, $1240eV.nm$for $hc$and $3.16\times {10}^{-3}eV$for $∆E$.

$$\begin{gathered} ∆\lambda =\frac{{(91.4nm)}^2}{1240eV.nm}(3.16\times {10}^{-3}eV) \\ =0.021nm \end{gathered}$$

therefore, the wavelength difference of two transitions is 0.021 nm.