Question

Suppose a hypothetical H-like atom produces a blue, yellow, red and violet line in the emission spectrum. Match the above lines with their corresponding possible electronic transition:

| Colour of spectral lines| | Possible corresponding transitions
---|---|---|---
a.| Blue| p.| $6 \to 3$
b.| Yellow| q.| $2 \to 1$
c.| Red| r.| $5 \to 2$
d.| Violet| s.| $4 \to 3$

The correct option is-

Answer 1

The line spectrum results from the emission of radiations from atoms of the elements and thus called the atomic spectrum whereas molecules give a band spectrum known as molecular spectrum. We know that energy carried by a light is directly proportional to its frequency, from this we can compare which lines produce which colours.

Complete Step by step answer:
Here, we just have to compare the energy transition of the spectrum to identify the colours produced by it. Before that let us understand what the emission spectrum is.
When an excited electron returns back from higher energy level to ground state. It emits energy in the form of radiations. The spectrum produced by these emitted radiation is known as emission spectrum.
The frequency of the spectral line is given as
$\nu = 3.29 \times {10^{15}}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$
Where ${n_1}$ and ${n_2}$ are the electronic levels involved in transition
Also we know that he energy is directly proportional to frequency hence we can say that
$Energy,E\; \propto frequency,\;\nu$
So we can relate the both equations as
$Energy\; \propto \left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$ Since the number are constant and we can ignore them

Now let us come back to the colours, we know that the colours in VIBGYOR are violet, indigo, blue, green, yellow, orange and red. There are many important details VIBGYOR can provide us such that the frequency of VIBGYOR increases from red to violet. So we can say red has the least frequency. Also the wavelength of VIBGYOR increases from violet to red.
As we said earlier energy is directly proportional to frequency. We can keep this in mind before moving to our next step.
(p)$6 \to 3$: Let's find out the energy emitted when this electronic transmission happens
Here, ${n_1} = 3$and ${n_2} = 6$
${E_{6 \to 3}} = {E_0}\left[ {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{6^2}}}} \right]$
Here we took a constant ${E_0}$ to remove the proportionality, it is just for comparison
${E_{6 \to 3}} = {E_0}\left[ {\dfrac{1}{9} - \dfrac{1}{{36}}} \right] = {E_0}\left[ {\dfrac{3}{{36}}} \right]$
${E_{6 \to 3}} = 0.0833{E_0}$
(q) $2 \to 1$
Here, ${n_1} = 1$and ${n_2} = 2$
${E_{2 \to 1}} = {E_0}\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right]$
${E_{2 \to 1}} = {E_0}\left[ {1 - \dfrac{1}{4}} \right] = {E_0}\left[ {\dfrac{3}{4}} \right]$
${E_{2 \to 1}} = 0.75{E_0}$
(r) $5 \to 2$
Here, ${n_1} = 2$and ${n_2} = 5$
${E_{5 \to 2}} = {E_0}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{5^2}}}} \right]$
${E_{5 \to 2}} = {E_0}\left[ {\dfrac{1}{4} - \dfrac{1}{{25}}} \right] = {E_0}\left[ {\dfrac{{21}}{{100}}} \right]$
${E_{5 \to 2}} = 0.21{E_0}$
(s) $4 \to 3$
Here, ${n_1} = 3$and ${n_2} = 4$
${E_{4 \to 3}} = {E_0}\left[ {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}} \right]$
${E_{4 \to 3}} = {E_0}\left[ {\dfrac{1}{9} - \dfrac{1}{{16}}} \right] = {E_0}\left[ {\dfrac{7}{{144}}} \right]$
${E_{4 \to 3}} = 0.0486{E_0}$
So the energy variations can be compared as
${E_{2 \to 1}} > {E_{5 \to 2}} > {E_{6 \to 3}} > {E_{4 \to 3}}$
${\text{q}} > {\text{r}} > {\text{s}} > {\text{p}}$
This is the order of energy
We have four colours blue, yellow, violet and red
Their order of energy will be as discussed above red has least energy and violet has highest energy and keeping VIBGYOR in mind we can compare the given colours
${\text{violet}} > {\text{blue}} > {\text{yellow}} > {\text{red}}$
Hence we say
q is violet (d)
r is blue (a)
p is yellow (b)
s is red (c)

Hence option (1) is correct.

Note: Here we took energy to compare them, we could have taken any measurement to compare them like frequency or wavelength. All of them will result in the same answer. For better understanding only we choose energy as the parameter.