The maximum velocity of the photoelectrons emitted by a metal surface is $9 \times 10^5 \ m/s$. The value of the ratio of charge $(e)$ to mass $(m)$ of the photoelectron is $1.8 \times 10^{11} \ C/kg$. The value of the stopping potential in volts is:

When a metal is irradiated by light having wavelength $\lambda$ $(\lambda < \lambda_0)$,all the photoelectrons emitted are bent in a circle of radius $r$ by a magnetic field of flux density $B_0$. Find $\frac{1}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.

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

Let $v_1$ and $v_2$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_1 = 4 \text{ eV}$ and $E_2 = 2.5 \text{ eV}$,respectively. If the work function of the metal is $2 \text{ eV}$,then the ratio $\frac{v_1}{v_2}$ is

$A$ 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:

According to the photoelectric effect,the plot of kinetic energy of the emitted photo-electrons from a metal versus the frequency of the incident radiation gives a straight line whose slope