$A$ metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V_0$. If the same surface is illuminated with radiation of wavelength $2 \lambda$,the stopping potential becomes $\frac{V_0}{4}$. The threshold wavelength for this metallic surface will be -

$A$ metal surface having work function '$W_{0}$' emits photoelectrons when photons of energy '$E$' are incident on it. The electron enters a uniform magnetic field '$B$' in a perpendicular direction and moves in a circular path of radius '$r$'. Then '$r$' is equal to (where '$m$' and '$e$' are the mass and charge of the electron,respectively).

Which theory of light can explain the photoelectric effect?

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:

Radiation $p$ with wavelength $6561 \; \mathring{A}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} \; T$. If the radius of the largest circular path followed by the electrons is $10 \; mm$,the work function of the metal is close to ............... $eV$.

In the photoelectric effect,the kinetic energy $(K.E.)$ of electrons emitted from the metal surface depends upon: