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Question: One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. T...

One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve this dissociation lies in:
A. Visible region
B. Infrared region
C. Ultraviolet region
D. Microwave region

Explanation

Solution

The energy required to dissociate the bond between carbon and oxygen in carbon monoxide is given to be 11ev.
This energy required to dissociate the bond will be provided by the electromagnetic radiation.
And as we know, that the energy of an electromagnetic radiation is given by Planck’s law:
E=hνE = h\nu
Where h is the Planck’s constant =6.626×10−34Js = 6.626 \times {10^{ - 34}}Js
And ν\nu is the frequency of the given electromagnetic radiation,
Put E = 11ev and find the frequency of radiation needed, and compare it with the frequency range of various regions given in the option.

Complete step by step answer:
We first convert the given value of energy in eV into joules:
Since 1eV=1.6×10−19J1.6 \times {10^{ - 19}}J,
∴21ev=1.6×10−19×11 J\therefore 21ev = 1.6 \times {10^{ - 19}} \times 11{\text{ }}J.

Now, as we know when a bond is formed between two atoms some amount of energy is released, and if we have to dissociate this bond then we have to provide the same amount of energy, and this energy is called bond energy.
This energy can be provided in the form of heat or electromagnetic radiations.
And from Planck’s law the energy of an electromagnetic radiation is given by:
E=hνE = h\nu
Where, h is Planck’s constant=6.626×10−34Js6.626 \times {10^{ - 34}}Js
And ν\nu is frequency of the electromagnetic radiation.

We have, energy E=1.6×10−19×11 J1.6 \times {10^{ - 19}} \times 11{\text{ J}}
Substituting in formula E=hνE = h\nu , we get frequency
ν=Eh ⇒ν=1.6×10−19×116.626×10−34 ⇒ν=2.656×1015Hz  \nu = \dfrac{E}{h} \\\ \Rightarrow \nu = \dfrac{{1.6 \times {{10}^{ - 19}} \times 11}}{{6.626 \times {{10}^{ - 34}}}} \\\ \Rightarrow \nu = 2.656 \times {10^{15}}Hz \\\
The resultant frequency belongs to Ultraviolet region of light (i.e. 8 × 1014Hz   to 3 × 1016Hz8{\text{ }} \times {\text{ }}{10^{14}}Hz\;{\text{ to }}3{\text{ }} \times {\text{ }}{10^{16}}Hz)

So, the correct answer is “Option C”.

Note:
Remember to convert all units into one common unit system.
The student should be aware of the various frequency ranges of light, most importantly of the visible range i.e. 4×10144 \times {10^{14}} to 8×10148 \times {10^{14}} Hz.