The photoelectric work function of a metal is $1 \ eV$. Light of wavelength $\lambda = 3000 \ \mathring{A}$ is incident on it. What will be the velocity of the emitted photoelectrons?

If the work function of a photometal is $6.825 \; eV$,its threshold wavelength will be ............. $\mathring A$ $(c = 3 \times 10^8 \; m/s)$.

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 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:

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630 \; \mathring{A}$ is $0.42 \; V$. If the threshold frequency is $x \times 10^{13} \; s^{-1}$,where $x$ is ... (nearest integer).
(Given: speed of light $= 3 \times 10^{8} \; m/s$,Planck's constant $= 6.63 \times 10^{-34} \; J \cdot s$)

When a metallic surface is illuminated with a radiation of wavelength $\lambda$, the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $6\lambda$, the stopping potential is $V/12$. The threshold wavelength for the surface is

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface,having work function $5.01 \, eV$,when ultraviolet light of $200 \, nm$ falls on it,must be ............... $V$.