According to Einstein's Photoelectric Equation:

Given that incident frequency is $\nu = 2\nu_0$:

Since some emitted electrons lose energy through internal collisions within the metal lattice before escaping, their actual kinetic energies vary continuously in the range $0 \le K \le K_{\max}$. Thus, the maximum kinetic energy of emitted photoelectrons can be $h\nu_0$.

• For a point source of light of power $P$, the intensity of radiation at distance $d$ follows the inverse square law:
  $$I = \frac{P}{4\pi d^2} \propto \frac{1}{d^2}$$
• Photoelectric current $i$ is directly proportional to the intensity of incident radiation (number of photons incident per unit area per second):
  $$i \propto I \implies i \propto \frac{1}{d^2}$$

The de Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through potential difference $V$ is:

When the potential difference is doubled ($V' = 2V$):

Since $\frac{\lambda}{\sqrt{2}} \lt \lambda$, the de Broglie wavelength decreases.