The threshold frequency is defined as the minimum frequency of the incident radiation below which the photoelectric emission is not possible completely. It is given as,
\(\nu_o=\frac{E_o}{h}\) (1)
Where,
ν₀=thereshold frequency
E₀=work function in joules
h=planks constant=6.6×10⁻³⁴ m²kg s⁻¹
The values given in the question are,
\(E_o=5 \times 1.6 \times 10^{-19}=8 \times 10^{-19} J\) (2)
By placing the value of E₀ and h in equation (1) we get;
\(\nu_o=\frac{8 \times 10^{-19}}{6.6 \times 10^{-34}}=1.21 \times 10^{15} Hz\)
Hence, the threshold frequency for the metal is 1.21×10¹⁵Hz.