Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

If the work function of a metal is $\phi$ and the frequency of the incident light is $\nu$,there is no emission of photoelectrons if:

When a piece of metal is illuminated by a monochromatic light of wavelength $\lambda$,the stopping potential is $3 V_{s}$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential becomes $V_{s}$. The value of the threshold wavelength for photoelectric emission is:

For a metal with a work function of $6.6 \text{eV}$, which of the following wavelengths of incident radiation does not cause the photoelectric effect (in $\text{nm}$)? (Take Planck's constant $h = 6.6 \times 10^{-34} \text{J s}$ and speed of light $c = 3 \times 10^8 \text{m/s}$)

The threshold wavelength of tungsten is $2300 \, \mathring{A}$. If ultraviolet light of wavelength $1800 \, \mathring{A}$ is incident on it,then the maximum kinetic energy of photoelectrons would be about ............ $eV$.

In a photoelectric experiment, the wavelength of the light incident on the metal is changed from $200 \, nm$ to $400 \, nm$. The decrease in the stopping potential is close to [Use $hc = 1240 \, eV \cdot nm$ where $h$ is Planck's constant and $c$ is the velocity of light]. (in $ \, V$)