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.

In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

Light of frequency $4v_0$ is incident on a metal surface with a threshold frequency $v_0$. The maximum kinetic energy of the emitted photoelectrons is:

This question has Statement $1$ and Statement $2.$ Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement $1:$ $A$ metallic surface is irradiated by a monochromatic light of frequency $v > v_0$ (the threshold frequency). If the incident frequency is now doubled,the photocurrent and the maximum kinetic energy are also doubled.
Statement $2:$ The maximum kinetic energy of photoelectrons emitted from a surface is linearly dependent on the frequency of the incident light. The photocurrent depends only on the intensity of the incident light.

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by a light of wavelength $\lambda_1$ and $K_2$ corresponding to $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

Five elements $A, B, C, D$ and $E$ have work functions $1.2 \, eV, 2.4 \, eV, 3.6 \, eV, 4.8 \, eV$ and $6 \, eV$ respectively. If light of wavelength $4000 \, Å$ is allowed to fall on these elements, then photoelectrons are emitted by: