Assuming photoemission to take place,the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times,is

Light of wavelength $\lambda_{\text{ph}}$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $\phi$ and the anode is a wire mesh of conducting material kept at a distance $d$ from the cathode. $A$ potential difference $V$ is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $\lambda_e$,which of the following statement$(s)$ is(are) true?

In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

An image of the sun is formed by a lens of focal length $30 \ cm$ on the metal surface of a photoelectric cell and a photoelectric current $I$ is produced. The lens forming the image is then replaced by another of the same diameter but of focal length $15 \ cm$. The photoelectric current in this case is

Two incident radiations having energies two times and ten times of the work function of a metal surface,produce photoelectric effect. The ratio of maximum velocities of emitted photoelectrons respectively is

In a photoelectric experiment,ultraviolet light of wavelength $280 \, nm$ is used with a lithium cathode having a work function $\phi = 2.5 \, eV$. If the wavelength of incident light is switched to $400 \, nm$,find out the change in the stopping potential (in $V$).
$(h = 6.63 \times 10^{-34} \, J \cdot s, c = 3 \times 10^8 \, m/s)$