In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of the photoelectrons will become:

In a photoelectric effect experiment,the cathode metal is exposed to light of wavelength $600 \ nm$. The maximum kinetic energy of the ejected electron doubles when light of wavelength $400 \ nm$ is used. The work function of the cathode metal is approximately: [Use $h=6.63 \times 10^{-34} \ J-s, c=3 \times 10^8 \ m/s$] (in $eV$)

In a photoelectric experiment,three different lights are incident on a metal with a work function of $1.5 \ eV$. Light $A$ has a wavelength of $200 \ nm$ and an intensity of $1.8 \ W/m^2$,light $B$ has a wavelength of $400 \ nm$ and an intensity of $1 \ W/m^2$,and light $C$ has a wavelength of $600 \ nm$ and an intensity of $0.5 \ W/m^2$. The photocurrent versus voltage is measured. Which graphs correspond to which light?

The work function of a metal is $h \nu_0$. Light of frequency $\nu$ falls on this metal. The photoelectric effect will take place only if

The surface of a certain metal is first illuminated with light of wavelength $\lambda_1 = 350 \ nm$ and then with light of wavelength $\lambda_2 = 540 \ nm$. It is found that the maximum speed of the photoelectrons in the two cases differs by a factor of $2$. The work function of the metal (in $eV$) is close to: (Energy of photon $= \frac{1240}{\lambda \text{ (in } nm)} \ eV$)

Maximum velocity of photoelectrons emitted by a metal surface is $1.2 \times 10^6\, m/s$. Assuming the specific charge of the electron to be $1.8 \times 10^{11}\, C/kg$,the value of the stopping potential in volt will be