When radiation of the wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then,the threshold wavelength for the surface is :

The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function $\phi$,when light of frequency $\nu$ is incident on it,is $\lambda$. Then $\nu =$ (where $h$ is Planck's constant and $m$ is the mass of the electron).

If the work function of $Na$ metal is $2.3 \ eV$,in which of the following regions of the $EM$ spectrum does the threshold wavelength lie?

Photoelectrons are emitted when photons of energy $4.2 \text{ eV}$ are incident on a photosensitive metallic sphere of radius $10 \text{ cm}$ and work function $2.4 \text{ eV}$. The number of photoelectrons emitted before the emission is stopped is
$\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI unit; } e=1.6 \times 10^{-19} \text{ C}\right]$

Two streams of photons,possessing energies equal to twice and ten times the work function of a metal,are incident on the metal surface successively. The ratio of the maximum velocities of the photoelectrons emitted in the two respective cases is $x:y$. The value of $x$ is ...... .

$A$ photosensitive metallic surface has a work function $\phi$. If a photon of energy $3 \phi$ falls on the surface,the electron is emitted with a maximum velocity of $6 \times 10^6 \ m/s$. When the photon energy is increased to $9 \phi$,the maximum velocity of the photoelectrons will be: