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 -

To find the threshold frequency of light for the photoelectric emission from the metal, we can use the equation of the photoelectric effect:

Kmax = h*(f - f0)

where:

• Kmax is the maximum kinetic energy of the ejected electrons, which is given as 3 eV. We need to convert this to Joules by multiplying by the charge of an electron (1.6 * 10^-19 C), so Kmax = 3 * 1.6 * 10^-19 J = 4.8 * 10^-19 J.
• h is Planck's constant, which is 6.63 * 10^-34 J*s.
• f is the frequency of the incident light, which is given as 1 * 10^15 Hz.
• f0 is the threshold frequency, which we are trying to find.

Rearranging the equation to solve for f0 gives:

f0 = f - Kmax/h

Substituting the given values gives:

f0 = 1 * 10^15 Hz - 4.8 * 10^-19 J / 6.63 * 10^-34 J*s = 1 * 10^15 Hz - 7.24 * 10^14 Hz = 2.76 * 10^14 Hz

So, the threshold frequency of light for the photoelectric emission from the metal is approximately 2.76 * 10^14 Hz.