When a certain photosensitive surface is illuminated with...

Using Einstein's photoelectric equation, the maximum kinetic energy is K_\max = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}, where $\lambda_0$ is the threshold wavelength.

For the first case, when the wavelength is $\lambda$ and maximum velocity is $v$:

$\dfrac{1}{2}mv^2 = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}$

For the second case, when the wavelength is $\dfrac{3\lambda}{4}$ and maximum velocity is $2v$:

$\dfrac{1}{2}m(2v)^2 = \dfrac{hc}{\left(\dfrac{3\lambda}{4}\right)} - \dfrac{hc}{\lambda_0}$

$4\left(\dfrac{1}{2}mv^2\right) = \dfrac{4hc}{3\lambda} - \dfrac{hc}{\lambda_0}$

Substituting the value of $\dfrac{1}{2}mv^2$ from the first equation into the second equation:

$4\left(\dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}\right) = \dfrac{4hc}{3\lambda} - \dfrac{hc}{\lambda_0}$

Dividing the entire equation by $hc$:

$\dfrac{4}{\lambda} - \dfrac{4}{\lambda_0} = \dfrac{4}{3\lambda} - \dfrac{1}{\lambda_0}$

Rearranging the terms:

$\dfrac{4}{\lambda} - \dfrac{4}{3\lambda} = \dfrac{4}{\lambda_0} - \dfrac{1}{\lambda_0}$

$\dfrac{12 - 4}{3\lambda} = \dfrac{3}{\lambda_0}$

$\dfrac{8}{3\lambda} = \dfrac{3}{\lambda_0}$

$\lambda_0 = \dfrac{9\lambda}{8}$