The work function of caesium is $2.14 \ eV$. Find the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 \ V$. (Answer in $nm$)

$A$ light wave described by $E = 60 \sin(3 \times 10^{15} t) + \sin(12 \times 10^{15} t)$ (in $SI$ units) falls on a metal surface of work function $2.8 \text{ eV}$. The maximum kinetic energy of the ejected photoelectron is (approximately) . . . . . . $\text{eV}$.

The graph of stopping potential $(V_{s})$ against frequency $(\nu)$ of incident radiation is plotted for two different metals '$P$' and '$Q$' as shown in the graph. If $\phi_{P}$ and $\phi_{Q}$ are the work functions of metals '$P$' and '$Q$' respectively,then which of the following is correct?

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 work functions of Silver and Sodium are $4.6 \ eV$ and $2.3 \ eV$,respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is :

The wavelengths of light incident on a photocell are $400 \ nm$ and $250 \ nm$. The velocities of the emitted photoelectrons are $v$ and $2v$ respectively. What is the work function of the metal?