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?

The variation of stopping potential $(V_0)$ as a function of the frequency $\nu \ (\times 10^{14} \ Hz)$ of the incident light for a metal is shown in the figure. The work function of the surface is $........... \ eV$.

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491\, nm$ is $0.710\, V$. When the incident wavelength is changed to a new value,the stopping potential is $1.43\, V$. The new wavelength is ....... $nm.$

If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photo-electrons would be (given $h = 6.63 \times 10^{-34} \ J \cdot s$):

The energy of a photon with wavelength $\lambda$ is $2 \ eV$. When it strikes a metal surface,the maximum velocity of the emitted photoelectrons is $v$. If the value of $\lambda$ is decreased by $25\%$ and the maximum velocity is doubled,the work function of the metal becomes ...... $eV$.

$A$ certain metallic surface is illuminated by monochromatic radiation of wavelength $\lambda$. The stopping potential for photoelectric current for this radiation is $3V_{0}$. If the same surface is illuminated with a radiation of wavelength $2\lambda$,the stopping potential is $V_{0}$. The threshold wavelength of this surface for the photoelectric effect is $n\lambda$. Find the value of $n$.