Two radiations of photon energies $1 \; eV$ and $2.5 \; eV$ successively illuminate a photosensitive metallic surface of work function $0.5 \; eV$. The ratio of the maximum speeds of the emitted electrons is

For two different photosensitive materials having work functions $\phi$ and $2 \phi$ respectively,which are illuminated with light of sufficient energy to emit electrons,if the graph of stopping potential $(V_s)$ versus frequency $(\nu)$ is drawn for these two materials,what is the ratio of the slopes of the graphs for these two materials?

Explain: "For photoelectric emission, the frequency of incident radiation should be greater than the threshold frequency."

When the light of frequency $2v_0$ (where $v_0$ is threshold frequency) is incident on a metal plate,the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5v_0$,the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is

The stopping potential $({V_0})$:

For a photosensitive material,the work function is $W_0$ and the stopping potential is $V$. What is the wavelength of the incident radiation? ($h=$ Planck's constant,$c=$ velocity of light,$e=$ electronic charge)