On a photosensitive material,when the frequency of incident radiation is increased by $20 \%$,the maximum kinetic energy of emitted photoelectrons increases from $0.4 \ eV$ to $0.7 \ eV$. The work function of the material is (in $eV$)

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is:

Radiation of wavelength $300 \ nm$ and intensity $100 \ W \ m^{-2}$ falls on the surface of a photosensitive material. If $2 \%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $2 \ cm^2$ of the surface per second is nearly:

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).

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 $0.60 \ V$.

The work function of a metallic surface is $5.01 \, eV$. The photo-electrons are emitted when light of wavelength $2000 \, \mathring{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is ............... $volt$ $[h = 4.14 \times 10^{-15} \, eV \cdot s]$