An experimental study of the photoelectric effect involves a metal of work function $\phi_0$. What is the smallest wavelength of the incident photon to photoemit an electron of mass $m$ which has the same de Broglie wavelength as that of the incident photon? [Given $h$ is the Planck's constant, $c$ is the speed of light, and $\phi_0 \ll mc^2$]

The energy of the incident photon is $E = \dfrac{hc}{\lambda}$.

The momentum of the incident photon is $p = \dfrac{h}{\lambda}$.

By the photoelectric equation, the maximum kinetic energy of the photoemitted electron is $K_{\max} = \dfrac{hc}{\lambda} - \phi_0$.

The de Broglie wavelength of the emitted electron is given to be equal to the wavelength of the incident photon, $\lambda$.

Thus, the momentum of the electron is $p_e = \dfrac{h}{\lambda}$.

Using the non-relativistic expression for kinetic energy, the kinetic energy of the electron is $K = \dfrac{p_e^2}{2m} = \dfrac{h^2}{2m\lambda^2}$.

For the electron to be emitted, its kinetic energy must be less than or equal to the maximum kinetic energy available:

$\dfrac{h^2}{2m\lambda^2} \le \dfrac{hc}{\lambda} - \phi_0$

Rearranging the inequality, we get a quadratic in terms of $\dfrac{1}{\lambda}$:

$\dfrac{h^2}{2m}\left(\dfrac{1}{\lambda}\right)^2 - hc\left(\dfrac{1}{\lambda}\right) + \phi_0 \le 0$

Solving the corresponding quadratic equation for $\dfrac{1}{\lambda}$:

$\dfrac{1}{\lambda} = \dfrac{hc \pm \sqrt{h^2c^2 - 4\left(\dfrac{h^2}{2m}\right)\phi_0}}{2\left(\dfrac{h^2}{2m}\right)}$

$\dfrac{1}{\lambda} = \dfrac{hc \pm h\sqrt{c^2 - \dfrac{2\phi_0}{m}}}{\dfrac{h^2}{m}}$

$\dfrac{1}{\lambda} = \dfrac{mc}{h}\left(1 \pm \sqrt{1 - \dfrac{2\phi_0}{mc^2}}\right)$

The inequality holds between the two roots. To find the smallest possible wavelength $\lambda_{\min}$, we must take the maximum possible value of $\dfrac{1}{\lambda}$, which corresponds to the positive sign:

$\dfrac{1}{\lambda_{\min}} = \dfrac{mc}{h}\left(1 + \sqrt{1 - \dfrac{2\phi_0}{mc^2}}\right)$

Taking the reciprocal yields the smallest wavelength:

$\lambda_{\min} = \dfrac{h}{mc}\left(1 + \sqrt{1 - \dfrac{2\phi_0}{mc^2}}\right)^{-1}$

Answer: $\dfrac{h}{mc}\left(1 + \sqrt{1 - \dfrac{2\phi_0}{mc^2}}\right)^{-1}$