In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

When $UV$ light of wavelength $300 \, nm$ is incident on a metal surface having a work function of $2.13 \, eV$, electron emission takes place. The stopping potential is: (Given $hc = 1240 \, eV \cdot nm$) (in $ \, V$)

If the frequency of incident light in a photoelectric experiment is doubled,then the stopping potential will

The stopping potential for photoelectrons:

The threshold wavelengths for photoelectric emission from two metals $A$ and $B$ are $400 \ nm$ and $800 \ nm$ respectively. The ratio of their work functions,$\phi_{A} : \phi_{B}$ is:

The threshold frequency of a metal is $f_0$. When light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of the photoelectrons is $v_1$. When the frequency of the incident radiation is increased to $5f_0$, the maximum velocity of the photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is