If in a photoelectric experiment,the wavelength of incident radiation is reduced from $6000 \, \mathring{A}$ to $4000 \, \mathring{A}$,then:

The graph between the maximum speed $v_{\text{max}}$ of a photoelectron and frequency $(\nu)$ of the incident radiation,in the photoelectric effect,is correctly represented by:

In the photoelectric effect, the kinetic energy of the electron emitted from the metal surface depends upon:

$A$ metal surface of work function $1.07 eV$ is irradiated with light of wavelength $332 nm$. The retarding potential required to stop the escape of photo-electrons is .............. $V$.

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2\lambda$,the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is:

An electron in a hydrogen atom jumps from the second Bohr orbit to the ground state,and the energy difference of the two states is radiated in the form of photons. These photons are then allowed to fall on a metal surface having a work function equal to $4.2 \ eV$. Calculate the stopping potential. [Energy of electron in $n^{\text{th}}$ orbit $= -\frac{13.6}{n^2} \ eV$] (in $V$)