$A$ photosensitive surface has work function $\phi$. If a photon of energy $3 \phi$ falls on this surface, the electron comes out with a maximum velocity of $4 \times 10^6 \,m/s$. When the photon energy is increased to $7 \phi$, the maximum velocity of the photoelectron will be:

When radiation of wavelength $\lambda$ is incident on a photoelectric cell,the maximum velocity of the emitted photoelectrons is $\upsilon$. If the wavelength of the incident radiation is changed to $3\lambda / 4$,the maximum velocity of the emitted photoelectrons will be ..........

The graph of stopping potential $(V_{s})$ against frequency $(\nu)$ of incident radiation is plotted for two different metals '$P$' and '$Q$' as shown in the graph. If $\phi_{P}$ and $\phi_{Q}$ are the work functions of metals '$P$' and '$Q$' respectively,then which of the following is correct?

Monochromatic radiation emitted when an electron in a hydrogen atom jumps from the first excited state to the ground state irradiates a photosensitive material. The stopping potential is measured to be $3.57 \; V$. The threshold frequency of the material is ......... $\times 10^{15} \; Hz$.

When light of frequency $v_{1}$ is incident on a metal with work function $W$ (where $h v_{1} > W$),then the photocurrent falls to zero at a stopping potential of $V_{1}$. If the frequency of light is increased to $v_{2}$,the stopping potential changes to $V_{2}$. Therefore,the charge of an electron $e$ is given by:

If the maximum kinetic energy of emitted electrons in the photoelectric effect is $3.2 \times 10^{-19} \text{ J}$ and the work function for the metal is $6.63 \times 10^{-19} \text{ J}$,then the stopping potential and threshold wavelength respectively are:
[Planck's constant $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$]
[Velocity of light $c = 3 \times 10^{8} \text{ m/s}$]
[Charge on electron $e = 1.6 \times 10^{-19} \text{ C}$]