In the case of photoelectric emission from a certain metal,the cutoff frequency is $v$. If radiation of frequency $2v$ is incident on the metal plate,the maximum possible velocity of the emitted electrons will be ($m=$ mass of electron).

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 $eV$)

$A$ photon of energy $3.4 \ eV$ is incident on a metal having work function $2 \ eV$. The maximum kinetic energy $(K.E.)$ of photo-electrons is equal to ........... $eV$. (in $eV$)

The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3v}{2}$. Then the work function of the metal must be

For photoelectric emission from a certain metal,the cutoff frequency is $\nu$. If radiation of frequency $2\nu$ impinges on the metal plate,the maximum possible velocity of the emitted electron will be ($m$ is the electron mass).

The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$ is (Where $h =$ Planck's constant,$m =$ mass of electron,and $c =$ speed of light).