One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in the
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → ultraviolet region
The radiation can dissociate the molecule only if each photon energy \( E=h\nu \) is at least 11 eV. So the minimum frequency is \( \nu=E/h \). Its wavelength \( \lambda=c/\nu \) then fixes the band of the electromagnetic spectrum.
Taking \( h=6.63\times10^{-34}\,\mathrm{J\,s} \), \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \) and \( 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J} \):
\( \begin{aligned} E &= 11\,\mathrm{eV} \\ &= 11\times1.6\times10^{-19}\,\mathrm{J} \\ &= 1.76\times10^{-18}\,\mathrm{J} \end{aligned} \)
\( \begin{aligned} \nu &= \frac{E}{h} \\ &= \frac{1.76\times10^{-18}\,\mathrm{J}}{6.63\times10^{-34}\,\mathrm{J\,s}} \\ &\approx 2.65\times10^{15}\,\mathrm{Hz} \end{aligned} \)
\( \begin{aligned} \lambda &= \frac{c}{\nu} \\ &= \frac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{2.65\times10^{15}\,\mathrm{Hz}} \\ &\approx 1.13\times10^{-7}\,\mathrm{m} \\ &= 113\,\mathrm{nm} \end{aligned} \)
From the above, \( \lambda\approx113\,\mathrm{nm} \). This is shorter than 400 nm, the violet end of visible light. It lies in the ultraviolet band, 400 nm down to 0.6 nm. Hence, the radiation lies in the ultraviolet region.
NCERT (2026-27):
Ultraviolet Rays – “It covers wavelengths ranging from about 4 × 10–7 m (400 nm) down to 6 × 10–10m (0.6 nm).”