Question

максимальная кинетическая энергия фотоэлектрона калия при его освещении светом с длинной волны 400 нм. если работа выхода электронов у калия равна 3,6*10 в минус 19 дж

Answer 1

Решение во вложении..!

Ответ: E(k) = 1,365*10⁻¹⁹ Дж.

 

Answer 2

Calculation of the Maximum Kinetic Energy of a Potassium Photoelectron

To calculate the maximum kinetic energy of a photoelectron, we can use the equation:

Kinetic Energy = Energy of the Incident Photon - Work Function

The energy of a photon can be calculated using the equation:

Energy of a Photon = Planck's Constant × Speed of Light / Wavelength

Given: - Wavelength of light = 400 nm = 400 × 10^-9 m - Work function of potassium = 3.6 × 10^-19 J

Let's calculate the maximum kinetic energy of the photoelectron.

Using the equation for the energy of a photon:

Energy of a Photon = (Planck's Constant × Speed of Light) / Wavelength

Substituting the values:

Energy of a Photon = (6.626 × 10^-34 J·s × 3 × 10^8 m/s) / (400 × 10^-9 m)

Calculating the energy of the photon gives us:

Energy of a Photon ≈ 4.969 × 10^-19 J

Now, we can calculate the maximum kinetic energy of the photoelectron using the equation:

Kinetic Energy = Energy of the Photon - Work Function

Substituting the values:

Kinetic Energy = (4.969 × 10^-19 J) - (3.6 × 10^-19 J)

Calculating the maximum kinetic energy gives us:

Kinetic Energy ≈ 1.369 × 10^-19 J

Therefore, the maximum kinetic energy of the photoelectron is approximately 1.369 × 10^-19 J when potassium is illuminated with light of a wavelength of 400 nm and the work function of potassium is 3.6 × 10^-19 J.

Please note that the above calculations are based on the given values and assumptions.