Question

A metal with a threshold frequency of 1.42×10^15 s−1 emits an electron with a velocity of 7.77×10^5 m/s when radiation of 1.87×10^15 s−1 strikes the metal. What is the metal's work function?

A) 2.5 eV
B) 4.0 eV
C) 1.2 eV
D) 3.3 eV

Answer 1

The work function of a metal is determined using the photoelectric effect equation, by subtracting the kinetic energy of an emitted electron from the energy of the incoming photon and expressing the result in electronvolts.

Step-by-step explanation:

The question involves the concept of the photoelectric effect in physics, where the work function is the minimum energy required to eject an electron from the surface of a metal.

To find the metal's work function, we use Einstein's photoelectric equation: KE = hf - φ, where KE is the kinetic energy of the emitted electron, hf is the energy of the incident photon (with h being Planck's constant and f the frequency of incident radiation), and φ (phi) is the work function of the metal.

The kinetic energy (KE) of the emitted electron can be found using the formula KE = 1/2 mv2, where m is the mass of the electron and v is its velocity. By substituting the given values and calculating the kinetic energy in joules, and then converting it to electronvolts (eV), we can solve for the work function. Since the threshold frequency of the metal is given, it can also be used to compute the energy hf of the incident photon. Subtracting the kinetic energy from the photon energy gives the work function in electronvolts, which can be matched with one of the provided options.