Final answer:
The maximum kinetic energy of electrons emitted when gold is illuminated with ultraviolet radiation of frequency 1.7 x 10^15 Hz is 3.38 x 10^-19 joules or 2.1125 electron volts (eV). The stopping potential for these electrons is 2.1125 volts.
Step-by-step explanation:
The question concerns the Photoelectric Effect, where light of a certain frequency can cause the emission of electrons from a metal surface. The work function represents the minimum energy needed to remove an electron from the surface of a metal. We'll answer this question using the equation for the Photoelectric Effect, which relates the work function, the energy of the incident photons, and the maximum kinetic energy of the emitted electrons.
Part (a): Calculating maximum kinetic energy in joules
We use the equation: KEmax = E - work function, where E is the energy of the incident photons and is given by E = h * f. Here, h is Planck's constant, and f is the frequency.
- E = (6.6 x 10-34 J s) * (1.7 x 1015 Hz)
- E = 1.122 x 10-18 J
- KEmax = E - (4.9 eV * 1.6 x 10-19 C/eV)
- KEmax = 1.122 x 10-18 J - 7.84 x 10-19 J
- KEmax = 3.38 x 10-19 J
Part (b): Expressing the energy in eV
To convert the kinetic energy from joules to eV, divide by the charge of an electron:
- KEmax (eV) = KEmax (J) / (1.6 x 10-19 C)
- KEmax (eV) = 3.38 x 10-19 J / (1.6 x 10-19 C/eV)
- KEmax (eV) = 2.1125 eV
Part (c): Calculating the stopping potential
The stopping potential Vs is related to KEmax as follows: eVs = KEmax. Solving for Vs:
- Vs = KEmax / e
- Vs = 3.38 x 10-19 J / (1.6 x 10-19 C)
- Vs = 2.1125 V