Answer
$\lambda = 3.64\times 10^{-12}~m$
Work Step by Step
We can find the minimum energy of the photon:
$E = m_e~c^2+m_p~c^2$
$E = (9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2+(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$
$E = 5.46\times 10^{-14}~J$
We can find the maximum wavelength of the photon:
$E = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(5.46\times 10^{-14}~J)}$
$\lambda = 3.64\times 10^{-12}~m$
