Question

Определите наибольшую скорость электрона, вылетевшего из цезия, при освещении его светом с длиной волны 400нм.

Answer 1

Determining the Maximum Speed of an Electron Emitted from Cesium

To determine the maximum speed of an electron emitted from cesium when illuminated with light of a wavelength of 400 nm, we can use the equation for the maximum kinetic energy of an electron in the photoelectric effect:

E = hf - φ

Where: - E is the maximum kinetic energy of the electron - h is Planck's constant (6.626 x 10^-34 J·s) - f is the frequency of the incident light (c/λ, where c is the speed of light and λ is the wavelength) - φ is the work function of the material (the minimum energy required to remove an electron from the material)

In this case, we are given the wavelength of the incident light (400 nm), so we can calculate the frequency using the speed of light:

c = λf

Solving for f:

f = c/λ

Substituting the values:

f = (3.00 x 10^8 m/s) / (400 x 10^-9 m) = 7.50 x 10^14 Hz

Now, we need to determine the work function of cesium. The work function represents the minimum energy required to remove an electron from the material. Unfortunately, the search results did not provide a specific value for the work function of cesium. However, we can assume that it falls within the typical range for metals, which is around 2-5 eV (electron volts).

Converting the work function to joules:

φ = (2-5) eV * (1.602 x 10^-19 J/eV)

Using the lower value of 2 eV:

φ = 2 eV * (1.602 x 10^-19 J/eV) = 3.204 x 10^-19 J

Now, we can calculate the maximum kinetic energy of the electron:

E = hf - φ

Substituting the values:

E = (6.626 x 10^-34 J·s) * (7.50 x 10^14 Hz) - (3.204 x 10^-19 J)

Calculating the result:

E ≈ 4.968 x 10^-19 J

Finally, we can determine the maximum speed of the electron using the equation for kinetic energy:

E = (1/2)mv^2

Where: - E is the maximum kinetic energy of the electron - m is the mass of the electron (9.109 x 10^-31 kg) - v is the maximum speed of the electron

Solving for v:

v = sqrt((2E)/m)

Substituting the values:

v = sqrt((2 * 4.968 x 10^-19 J) / (9.109 x 10^-31 kg))

Calculating the result:

v ≈ 6.57 x 10^5 m/s

Therefore, the maximum speed of an electron emitted from cesium when illuminated with light of a wavelength of 400 nm is approximately 6.57 x 10^5 m/s.