What is the maximum wavelength of electromagnetic radiation that could be used to photodisintegrate a deuteron?

$\lambda_{\mathrm{max}}=5.57\times 10^{-13}\ \mathrm{m}$

The binding energy of the deuteron is $2.23\ \mathrm{MeV}$, so a photon would need to have at least this much energy to break the deuteron into a proton and neutron. Since

$$E=hf$$

and

$$f=\displaystyle \frac{c}{\lambda }$$

$$\Rightarrow $$

$$E=\frac{hc}{\lambda }$$

$$\lambda_{\mathrm{max}}=\frac{hc}{E_{\mathrm{min}}}$$

$$\lambda_{\mathrm{max}}=\frac{\left(4.14\times 10^{-15}\ \mathrm{eV\cdot s}\right)\left(3.00\times 10^8\ \mathrm{\displaystyle \frac{m}{s}}\right)}{2.23\times 10^6\ \mathrm{eV}}$$

$$\lambda_{\mathrm{max}}=5.57\times 10^{-13}\ \mathrm{m}$$