A hydrogen atom in an n = 2 state absorbs a photon,

1. What should be the photon wavelength to cause the electron to jump to an n = 4 state?
2. What wavelength photons might be emitted by the atom following this absorption?

Energy levels or energy state is any discrete value from a set of total energy values for a force-limited subatomic particle in a limited space, or for a system of such particles, such as an atom or a nucleus.

The energy of a state is defined by the principal quantum number and the difference in energy between two states is said to be the energy difference between two states.

As you know that energy of a state is

${\mathrm{E}}_{\mathrm{n}}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}\left(\mathrm{n}=1,2,3,.......\right)$

The energy difference between orbit and orbit 1 is given by,

role="math" localid="1659326526082" $$\begin{gathered} ∆\mathrm{E}=\mathrm{The}\mathrm{Energy}\mathrm{orbit}4-\mathrm{The}\mathrm{Energy}\mathrm{of}\mathrm{orbit}1\backslash \\ ={\mathrm{E}}_4-{\mathrm{E}}_1 \\ =\left(-13.6\mathrm{eV}\right)\left(\frac{1}{4^2}-\frac{1}{2^2}\right) \\ ∆\mathrm{E}=2.55\mathrm{eV} \end{gathered}$$

….. (1)

For calculating the wavelength, you will use the equation given below

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (2)

Here, $\lambda$is the wavelength, E is the energy of photon, h is the Planck’s constant, and c is the speed of light.

The Plank’s constant, hc = 1240 eV.nm

Substitute 2.55eV for E and 1240 eV. nm for hc into equation (2).

role="math" localid="1659326706949" $$\begin{gathered} \frac{1240\mathrm{eV}.\mathrm{nm}}{\mathrm{\lambda }}=2.55\mathrm{eV} \\ \mathrm{\lambda }=486\mathrm{nm} \end{gathered}$$

The electron can jump from n = 4 to n = 1 in the following different paths:

1. $4\to 2\mathrm{then}2\to 1$

2. $4\to 3\mathrm{then}3\to 2\mathrm{then}2\to 1$

3. $4\to 1$

4.$4\to 3\mathrm{then}3\to 1$

Hence, the emitted wavelengths will be corresponding to the above-mentioned jumps.

From step 3, the unique combination of jumps is,

$4\to 2,2\to 1,4\to 3,3\to 2,4\to 1\mathrm{and}3\to 1$.

Hence, wavelength corresponding to only these 6 jumps can be emitted.

As known that,

$\frac{1}{\mathrm{\lambda }}=1.097\times {10}^7{\mathrm{m}}^{-1}\left(\frac{1}{{\mathrm{n}}_{\mathrm{f}}^2}-\frac{1}{{\mathrm{n}}_{\mathrm{i}}^2}\right)$

Where, $\lambda$is the Wavelength emitted, ${\mathrm{n}}_{\mathrm{f}}$ is the final orbit, and ${\mathrm{n}}_{\mathrm{i}}$is the initial orbit.

Now, for$4\to 2$:

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7{\mathrm{m}}^{-1}\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(4\right)}^2}\right) \\ =1.097\times {10}^7{\mathrm{m}}^{-1}\left(0.25-0.0625\right) \\ =0.02056875\times {10}^7{\mathrm{m}}^{-1} \end{gathered}$$

$$\begin{gathered} \lambda =\frac{1}{0.02056875\times {10}^7{\mathrm{m}}^{-1}} \\ =4.86\times {10}^7\mathrm{m} \\ =4.86\times {10}^{-9}\mathrm{m} \\ =486\mathrm{nm} \end{gathered}$$

For $4\to 3$:

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7{\mathrm{m}}^{-1}\left(\frac{1}{{\left(3\right)}^2}-\frac{1}{{\left(4\right)}^2}\right) \\ =1.097\times {10}^7{\mathrm{m}}^{-1}\left(0.111-0.0625\right) \\ =0.0532045\times {10}^7{\mathrm{m}}^{-1} \end{gathered}$$

$$\begin{gathered} \mathrm{\lambda }=\frac{1}{0.0532\times {10}^7{\mathrm{m}}^{-1}} \\ =18.75\times {10}^7\mathrm{m} \\ =18.75\times {10}^{-9}\mathrm{m} \\ =1875\mathrm{nm} \end{gathered}$$

For $4\to 1$:

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7{\mathrm{m}}^{-1}\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(4\right)}^2}\right) \\ =1.097\times {10}^7{\mathrm{m}}^{-1}\left(1-0.0625\right) \\ =1.02843\times {10}^7{\mathrm{m}}^{-1} \end{gathered}$$

$$\begin{gathered} \mathrm{\lambda }=\frac{1}{1.02843\times {10}^7{\mathrm{m}}^{-1}} \\ =0.972\times {10}^7\mathrm{m} \\ =0.972\times {10}^{-9}\mathrm{m} \\ =97.2\mathrm{nm} \end{gathered}$$

For $2\to 1$:

localid="1659327531207" $$\begin{gathered} \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7{\mathrm{m}}^{-1}\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right) \\ =1.097\times {10}^7{\mathrm{m}}^{-1}\left(1-0.025\right) \\ =0.82275\times {10}^7{\mathrm{m}}^{-1} \\ \mathrm{\lambda }=\frac{1}{0.82275\times {10}^7{\mathrm{m}}^{-1}} \\ =1.22\times {10}^7\mathrm{m} \\ =122\times {10}^{-9}\mathrm{m} \\ =122\mathrm{nm} \end{gathered}$$

For $3\to 2$:

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7{\mathrm{m}}^{-1}\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(3\right)}^2}\right) \\ =1.097\times {10}^7{\mathrm{m}}^{-1}\left(0.25-0.111\right) \\ =0.1524\times {10}^7{\mathrm{m}}^{-1} \\ \mathrm{\lambda }=\frac{1}{0.1524\times {10}^7{\mathrm{m}}^{-1}} \\ =6.56\times {10}^7\mathrm{m} \\ =656\times {10}^{-9}\mathrm{m} \\ =656\mathrm{nm} \end{gathered}$$

For $3\to 1$:

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7{\mathrm{m}}^{-1}\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(3\right)}^2}\right) \\ =1.097\times {10}^7{\mathrm{m}}^{-1}\left(1-0.111\right) \\ =0.97523\times {10}^7{\mathrm{m}}^{-1} \end{gathered}$$

$$\begin{gathered} \mathrm{\lambda }=\frac{1}{0.97523\times {10}^7{\mathrm{m}}^{-1}} \\ =1.03\times {10}^{-7}\mathrm{m} \\ =103\times {10}^{-9}\mathrm{m} \\ =103\mathrm{nm} \end{gathered}$$

Hence, the possible wavelengths emitted are as below.

$\mathrm{\lambda }=486\mathrm{nm},103\mathrm{nm},656\mathrm{nm},122\mathrm{nm},1875\mathrm{nm},97.2\mathrm{nm}$