Questions: Part 1 (1 point) Which metal emits electrons with the greater velocity? Part 2 (1 point) What is the velocity of the electrons from potassium? m / s Part 3 (1 point) What is the velocity of the electrons from sodium? m / s

Solution Steps

We need to determine the velocities of electrons emitted from potassium and sodium when they are exposed to light. This involves understanding the photoelectric effect, where electrons are emitted from a material when it absorbs light.

The kinetic energy of the emitted electrons can be calculated using the photoelectric effect equation: \[ K.E. = h\nu - \phi \] where:

• \( K.E. \) is the kinetic energy of the emitted electrons,
• \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
• \( \nu \) is the frequency of the incident light,
• \( \phi \) is the work function of the metal.

The kinetic energy is also related to the velocity \( v \) of the electrons by: \[ K.E. = \frac{1}{2}mv^2 \] where \( m \) is the mass of an electron (\(9.109 \times 10^{-31} \, \text{kg}\)).

Rearranging for \( v \): \[ v = \sqrt{\frac{2K.E.}{m}} \]

Potassium and sodium have different work functions:

• Potassium (\( \phi_{\text{K}} \)) = 2.30 eV
• Sodium (\( \phi_{\text{Na}} \)) = 2.75 eV

Given the same incident light frequency, the metal with the lower work function will emit electrons with greater kinetic energy and thus greater velocity.

Since potassium has a lower work function (2.30 eV) compared to sodium (2.75 eV), electrons emitted from potassium will have greater kinetic energy and thus greater velocity.

Assuming the incident light has enough energy to overcome the work function, we can calculate the velocity using the given work function and the photoelectric effect equation.

Similarly, we calculate the velocity for sodium using its work function.

Final Answer