Einstein's photoelectric equation is given by .........

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

Assertion : When ultraviolet light is incident on a photocell,its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X-$ rays,both $V_0$ and $K_{max}$ increase.
Reason : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

$A$ beam of photons with an energy of $10.5 \ eV$ strikes a metal plate. The photoelectrons are emitted with a maximum velocity of $1.6 \times 10^6 \ m \ s^{-1}$. The work function of the metal is (Assume mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$). (in $eV$)

What is the stopping potential when a metal with a work function of $0.6 \ eV$ is illuminated with light of $2 \ eV$ (in $V$)?

If the frequency of incident radiation $(v)$ is increased,keeping other factors constant,the stopping potential ($v > v_0$,threshold frequency) will: