Kmax(in eV)=3, and frequency of light (inHz)=1×1015 for a metal used as a cathode in a photoelectric experiment. The threshold frequency of light for the photoelectric emission from the metal is -1×1014 Hz

It seems like there's a misunderstanding. The threshold frequency of light for the photoelectric emission from a metal cannot be negative. The threshold frequency is the minimum frequency of light required to eject an electron from a metal surface. It's a positive value.

However, if you want to find the work function (ϕ) of the metal used in the photoelectric experiment, you can use the equation of the photoelectric effect:

Kmax = hf - ϕ

where:

• Kmax is the maximum kinetic energy of the ejected electrons, which is given as 3 eV. But we need to convert it to Joules by multiplying by the charge of an electron (1.6 x 10^-19 C). So, Kmax = 3 eV * 1.6 x 10^-19 C/eV = 4.8 x 10^-19 J.
• h is Planck's constant (6.63 x 10^-34 J.s).
• f is the frequency of the incident light, which is given as 1 x 10^15 Hz.

Rearranging the equation to find ϕ gives:

ϕ = hf - Kmax

Substituting the given values into the equation gives:

ϕ = (6.63 x 10^-34 J.s * 1 x 10^15 Hz) - 4.8 x 10^-19 J

Calculate the above expression to find the work function ϕ.