When a certain metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $\frac{V}{3}$. The threshold wavelength for the surface is:

$A$ light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of $2 \ \text{eV}$. If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of the metal is $1 \ \text{eV}$). (in $\text{eV}$)

The work functions for tungsten and sodium are $4.5 \text{ eV}$ and $2.3 \text{ eV}$ respectively. If the threshold wavelength $\lambda_0$ for sodium is $5460 \text{ \AA}$, the value of $\lambda_0$ for tungsten is ............ $\text{ \AA}$.

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).

Light of wavelength '$\lambda$' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with some velocity, then the stopping potential:

The magnetic field associated with a light wave is given,at the origin,by $B = B_0 [\sin(3.14 \times 10^7 ct) + \sin(6.28 \times 10^7 ct)]$. If this light falls on a silver plate having a work function of $4.7 \ eV$,what will be the maximum kinetic energy of the photoelectrons? (in $eV$)