When a metal surface is illuminated by light of wavelength $\lambda$,the stopping potential is $8 \ V$. When the same surface is illuminated by light of wavelength $3 \lambda$,the stopping potential is $2 \ V$. The threshold wavelength for this surface is:

When ultraviolet radiation of a certain frequency falls on a potassium target,the photoelectrons released can be stopped completely by a retarding potential of $0.6 \, V$. If the frequency of the radiation is increased by $10 \%$,this stopping potential rises to $0.9 \, V$. The work function of potassium is ........ $eV$.

The maximum velocity of the photoelectron emitted by the metal surface is $V$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

Light of wavelength $\lambda_{\text{ph}}$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $\phi$ and the anode is a wire mesh of conducting material kept at a distance $d$ from the cathode. $A$ potential difference $V$ is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $\lambda_e$,which of the following statement$(s)$ is(are) true?

In the photoelectric effect,the kinetic energy $(K.E.)$ of electrons emitted from the metal surface depends upon:

When a photon of wavelength ${\lambda _1}$ is incident on a metal surface,the speed of the ejected fastest electron is $v_1 \ m/s$. If a photon of wavelength ${\lambda _2}$ is incident on the same metal,the speed of the ejected fastest electron is $v_2 \ m/s$. The value of $v_2^2 - v_1^2$ will be