Photoelectric emission takes place from a certain metal at threshold frequency $v$. If the radiation of frequency $4 v$ is incident on the metal plate, the maximum velocity of the emitted photoelectrons will be ( $\mathrm{m}=$ mass of photoelectron, $\mathrm{h}=$ Planck's constant)

According to Einstein’s photoelectric equation, $hf = \phi + K_{\text{max}}$, where $h$ is Planck’s constant, $f$ is the incident frequency, $\phi$ is the work function, and $K_{\text{max}}$ is the maximum kinetic energy of ejected electrons.

Given the threshold frequency $v$, the work function is $\phi = hv$.

For incident frequency $4v$, the photoelectric equation becomes $h(4v) = hv + K_{\text{max}}$. Solving yields $K_{\text{max}} = 3hv$.

Since $K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2$, equating gives $\frac{1}{2} m v_{\text{max}}^2 = 3hv$.

Solving for $v_{\text{max}}$, we find $v_{\text{max}}^2 = \frac{6hv}{m}$, so $v_{\text{max}} = \sqrt{\frac{6hv}{m}}$, which corresponds to option A.