Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength $6800 \, \mathring{A}$. Calculate threshold frequency $(v_{0})$ and work function $(W_{0})$ of the metal.

(N/A) Threshold wavelength of radiation $(\lambda_{0}) = 6800 \, \mathring{A} = 6800 \times 10^{-10} \, m$.
Threshold frequency $(v_{0})$ is given by the formula $v_{0} = \frac{c}{\lambda_{0}}$.
Substituting the values: $v_{0} = \frac{3 \times 10^{8} \, m s^{-1}}{6.8 \times 10^{-7} \, m} = 4.41 \times 10^{14} \, s^{-1}$.
Thus,the threshold frequency $(v_{0})$ of the metal is $4.41 \times 10^{14} \, s^{-1}$.
Work function $(W_{0})$ is given by $W_{0} = h v_{0}$.
Substituting the values: $W_{0} = (6.626 \times 10^{-34} \, J s)(4.41 \times 10^{14} \, s^{-1}) = 2.922 \times 10^{-19} \, J$.