$A$ beam of electromagnetic radiation of intensity $6.4 \times 10^{-5} \; W/cm^{2}$ is comprised of wavelength $\lambda = 310 \; nm$. It falls normally on a metal surface (work function $\varphi = 2 \; eV$) of surface area $1 \; cm^{2}$. If one in $10^{3}$ photons ejects an electron, the total number of electrons ejected in $1 \; s$ is $10^{x}$. Then $x$ is: $(hc = 1240 \; eV \cdot nm, 1 \; eV = 1.6 \times 10^{-19} \; J)$

In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength $(\lambda)$ of incident light and the corresponding stopping potential $(V_0)$ are given below:

$\lambda (\mu m)$	$V_0$ (Volt)
$0.3$	$2.0$
$0.4$	$1.0$
$0.5$	$0.4$

Given that $c = 3 \times 10^8 \ m \ s^{-1}$ and $e = 1.6 \times 10^{-19} \ C$, Planck's constant (in units of $J \ s$) found from such an experiment is:

The maximum kinetic energy of a photoelectron is $E$ when the wavelength of incident radiation is $\lambda$. If the wavelength of the incident radiation is reduced to $\frac{\lambda}{3}$,the maximum kinetic energy becomes $4E$. The work function of the metal is:

In a photoelectric experiment for $4000 \, \mathring{A}$ incident radiation,the potential difference to stop the ejection is $2 \, V$. If the incident light is changed to $3000 \, \mathring{A}$,then the potential required to stop the ejection of electrons will be

The work functions for metals $A, B$ and $C$ are $1.92 eV, 2.0 eV$ and $5 eV$ respectively. According to Einstein's photoelectric equation,which of these metals will emit photoelectrons for an incident radiation of wavelength $4100 \mathring{A}$?