The work function of a metal is $2.1 \text{ eV}$. Which of the following wavelengths will be able to emit photoelectrons from its surface?

The stopping potential for photoelectrons:

$A$ and $B$ are two metals with threshold frequencies $1.8 \times 10^{14} \ Hz$ and $2.2 \times 10^{14} \ Hz$. Two identical photons of energy $0.825 \ eV$ each are incident on them. Then photoelectrons are emitted by (Take $h = 6.6 \times 10^{-34} \ J \cdot s$)

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

Initially,a photon of wavelength $\lambda_1$ falls on a photocathode and emits an electron of maximum energy $E_1$. If the wavelength of the incident photon is changed to $\lambda_2$,the maximum energy of the electron emitted becomes $E_2$. Then the value of $hc$ ($h=$ Planck's constant,$c=$ velocity of light) is

When a metallic surface is illuminated with light of wavelength $\lambda$,the stopping potential is $x \ V$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $\frac{x}{3} \ V$. The threshold wavelength for the metallic surface is ..........