Question

It takes to break an carbon-chlorine single bond. Calculate the maximum wavelength of light for which an carbon-chlorine single bond could be broken by absorbing a single photon. Round your answer to significant digits.

Answer 1

It takes approximately
`5.43* 10^(-19)\; \rm J` of energy to break one
`\rm C-Cl` single bond.

The maximum wavelength of a photon that can break one such bond is approximately
`3.66* 10^(-7)\; \rm m` (in vacuum.) That's the same as
`3.66 * 10^(2)\; \rm nm` (rounded to three significant figures.)

Step-by-step explanation:

Energy per bond

The standard bond enthalpy of
`\rm C-Cl` single bonds is approximately
`\rm 327\; \rm kJ \cdot mol^(-1)` (note that the exact value can varies across sources.) In other words, it would take approximately
`327\; \rm kJ` of energy to break one mole of these bonds.

The Avogadro Constant
`N_A \approx 6.023* 10^(23)\; \rm mol^(-1)` gives the number of
`\rm C-Cl` bonds in one mole of these bonds. Based on these information, calculate the energy of one such bond:

`\begin{aligned}& E(\text{one $\mathrm{C-Cl}$ bond}) \\ &= \frac{E(\text{one mole of $\mathrm{C-Cl}$ bonds})}{N_A} = (327\; \rm kJ\cdot mol^(-1))/(6.023* 10^(23)\; \rm mol^(-1)) \\ &\approx 5.429* 10^(-22)\; \rm kJ = 5.429* 10^(-19)\; \rm J \end{aligned}`.

Therefore, it would take approximately
`5.43* 10^(-19)\; \rm J` of energy to break one
`\rm C-Cl` single bond.

Minimum frequency and maximum wavelength

The Einstein-Planck Relation relates the frequency
`f` of a photon to its energy
`E`:

`E = h \cdot f`.

The
`h` here represents the Planck Constant:

`h \approx 6.63 * 10^(-34)\; \rm J \cdot s`.

A photon that can break one
`\rm C-Cl` single bond should have more than
`5.43* 10^(-19)\; \rm J` of energy. Apply the Einstein-Planck Relation to find the frequency of a photon with exactly that much energy:

`\begin{aligned}f &= (E)/(h)\\ &\approx (5.43* 10^(-19)\; \rm J)/(6.63 * 10^(-34)\; \rm J\cdot s) \\ &\approx 8.19 * 10^(14)\; \rm s^(-1) = 8.19 * 10^(14)\; \rm Hz\end{aligned}`.

What would be the wavelength
`\lambda` of a photon with a frequency of approximately
`8.19 * 10^(14)\; \rm Hz`? 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:

`\displaystyle \lambda = \frac{(\text{speed of light})}{f}`.

In vacuum, the speed of light is
`c \approx 2.998* 10^(8)\; \rm m \cdot s^(-1)`. Therefore, the wavelength of that
`8.19 * 10^(14)\; \rm Hz` photon in vacuum would be:

`\begin{aligned} \lambda &= (c)/(f) \\ & \approx (2.998* 10^(8)\; \rm m \cdot s^(-1))/(8.19* 10^(14)\; \rm s^(-1)) \\ &\approx 3.66 * 10^(-7)\; \rm m = 3.66 * 10^(2)\; \rm nm\end{aligned}`.

(Side note: that wavelength corresponds to a photon in the ultraviolet region of the electromagnetic spectrum.)