Radiation of frequency 2ν₀ is incident on a metal with threshold frequency ν₀.The correct statement of the following is _______.

The correct option is: Maximum kinetic energy of photo electrons emitted can be hν₀

Step-by-step Explanation:

According to Einstein's photoelectric equation, the maximum kinetic energy ($K_{\max }$) of the emitted photoelectrons is given by the relation:
$K_{\max }=E-\Phi$

where:
- $E$ is the energy of the incident radiation.
- $\Phi$ is the work function (threshold energy) of the metal.

The energy of the incident radiation with frequency $\nu =2{\nu }_0$ is:
$E=h\nu =h\left(2{\nu }_0\right)=2h{\nu }_0$

The work function of the metal with threshold frequency ${\nu }_0$ is:
$\Phi =h{\nu }_0$

Substituting these values into Einstein's photoelectric equation, we get the maximum kinetic energy as:
$K_{\max }=2h{\nu }_0-h{\nu }_0=h{\nu }_0$

Since photoelectrons can undergo collisions and lose energy before escaping the metal surface, they can be emitted with any kinetic energy ranging from zero up to the maximum limit of $h{\nu }_0$. Therefore, the maximum kinetic energy of the emitted photoelectrons can be $h{\nu }_0$.