Respuesta :
Answer:
It takes approximately [tex]5.43\times 10^{-19}\; \rm J[/tex] of energy to break one [tex]\rm C-Cl[/tex] single bond.
The maximum wavelength of a photon that can break one such bond is approximately [tex]3.66\times 10^{-7}\; \rm m[/tex] (in vacuum.) That's the same as [tex]3.66 \times 10^{2}\; \rm nm[/tex] (rounded to three significant figures.)
Explanation:
Energy per bond
The standard bond enthalpy of [tex]\rm C-Cl[/tex] single bonds is approximately [tex]\rm 327\; \rm kJ \cdot mol^{-1}[/tex] (note that the exact value can varies across sources.) In other words, it would take approximately [tex]327\; \rm kJ[/tex] of energy to break one mole of these bonds.
The Avogadro Constant [tex]N_A \approx 6.023\times 10^{23}\; \rm mol^{-1}[/tex] gives the number of [tex]\rm C-Cl[/tex] bonds in one mole of these bonds. Based on these information, calculate the energy of one such bond:
[tex]\begin{aligned}& E(\text{one $\mathrm{C-Cl}$ bond}) \\ &= \frac{E(\text{one mole of $\mathrm{C-Cl}$ bonds})}{N_A} = \frac{327\; \rm kJ\cdot mol^{-1}}{6.023\times 10^{23}\; \rm mol^{-1}} \\ &\approx 5.429\times 10^{-22}\; \rm kJ = 5.429\times 10^{-19}\; \rm J \end{aligned}[/tex].
Therefore, it would take approximately [tex]5.43\times 10^{-19}\; \rm J[/tex] of energy to break one [tex]\rm C-Cl[/tex] single bond.
Minimum frequency and maximum wavelength
The Einstein-Planck Relation relates the frequency [tex]f[/tex] of a photon to its energy [tex]E[/tex]:
[tex]E = h \cdot f[/tex].
The [tex]h[/tex] here represents the Planck Constant:
[tex]h \approx 6.63 \times 10^{-34}\; \rm J \cdot s[/tex].
A photon that can break one [tex]\rm C-Cl[/tex] single bond should have more than [tex]5.43\times 10^{-19}\; \rm J[/tex] of energy. Apply the Einstein-Planck Relation to find the frequency of a photon with exactly that much energy:
[tex]\begin{aligned}f &= \frac{E}{h}\\ &\approx \frac{5.43\times 10^{-19}\; \rm J}{6.63 \times 10^{-34}\; \rm J\cdot s} \\ &\approx 8.19 \times 10^{14}\; \rm s^{-1} = 8.19 \times 10^{14}\; \rm Hz\end{aligned}[/tex].
What would be the wavelength [tex]\lambda[/tex] of a photon with a frequency of approximately [tex]8.19 \times 10^{14}\; \rm Hz[/tex]? The exact answer to that depends on the medium that this photon is travelling through. To be precise, the exact answer depends on the speed of light in that medium:
[tex]\displaystyle \lambda = \frac{(\text{speed of light})}{f}[/tex].
In vacuum, the speed of light is [tex]c \approx 2.998\times 10^{8}\; \rm m \cdot s^{-1}[/tex]. Therefore, the wavelength of that [tex]8.19 \times 10^{14}\; \rm Hz[/tex] photon in vacuum would be:
[tex]\begin{aligned} \lambda &= \frac{c}{f} \\ & \approx \frac{2.998\times 10^{8}\; \rm m \cdot s^{-1}}{8.19\times 10^{14}\; \rm s^{-1}} \\ &\approx 3.66 \times 10^{-7}\; \rm m = 3.66 \times 10^{2}\; \rm nm\end{aligned}[/tex].
(Side note: that wavelength corresponds to a photon in the ultraviolet region of the electromagnetic spectrum.)
