If ultraviolet radiation of $6.2 eV$ falls on an aluminium surface,then the kinetic energy of the fastest emitted electron is (work-function $= 4.2 eV$)

In an experiment on the photoelectric effect,the wavelength of the incident radiation is $\lambda$. The wavelength of the incident radiation is reduced to $\frac{1}{3}$ of the initial value,and the maximum kinetic energy of the photoelectron is observed to be $n$ times the previous value. The threshold wavelength for the metal plate is:

In a photoelectric experiment,electrons are ejected from metals $X$ and $Y$ by light of intensity $I$ and frequency $f$. The potential difference $V$ required to stop the electrons is measured for various frequencies. If $Y$ has a greater work function than $X$,which one of the following graphs best illustrates the expected results?

Radiations of two photons having energies twice and five times the work function of a metal are incident successively on a metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases will be

Two light waves of wavelengths $600 \,nm$ and $200 \,nm$ are incident on a metal surface. The maximum velocity of photoelectrons produced due to one wavelength is $\frac{1}{3}$ of the maximum velocity of the photoelectrons produced due to the other wavelength. The work function of the metal is:

The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is $1.65 \ eV$) will be