Question

Наибольшая длина волны света, при которой ещё может наблюдаться фотоэффект на калии, равна 430 мкм. Найти скорость электронов, вырванных из калия светом с длиной волны 3 ∙10–7 м. Ответ выразить в км/с. Необходимо подробное решение, ответ я и сам могу найти.

Answer 1

Ответ:

Объяснение:

Дано:

λкр = 430 мкм = 430·10⁻⁶ м - красная граница для калия

λ = 3·10⁻⁷ м

______________

V - ?

1)

Красная граница для калия:

λкр = h·c / Авых

Работа выхода:

Aвых = h·c / λкр

2)

Энергия фотона:

ε = h·c / λ

Запишем уравнение Эйнштейна для фотоэффекта:

ε = Aвых + W

Кинетическая энергия:

W = ε - Aвых = h·c / λ - h·c / λкр =

= h·c·( 1/λ - 1/λкр) = 6,63·10⁻³⁴·3·10⁸·(1/(3·10⁻⁷) - 1/(430·10⁻⁶) ) ≈ 6,6·10⁻¹⁹ Дж

3)

Но

W = m·V²/2

Скорость фотоэлектронов:

V = √ (2·W / m) = √ (2·6,6·10⁻¹⁹/(9,1·10⁻³¹)) ≈ 1,2·10⁶ м/с  или

V = 1 200 км/с

Answer 2

Photoelectric Effect and Electron Speed Calculation

The maximum wavelength of light for which the photoelectric effect can still be observed on potassium is given as 430 µm. We are asked to find the speed of electrons ejected from potassium by light with a wavelength of 3 × 10⁻⁷ m, and express the answer in km/s.

To calculate the speed of the electrons, we can use the formula for the maximum kinetic energy of the emitted electrons in the photoelectric effect:

E = hf - Φ

Where: - E = maximum kinetic energy of the emitted electrons - h = Planck's constant (6.626 × 10⁻³⁴ J·s) - f = frequency of the incident light - Φ = work function of the metal (energy required to remove an electron from the metal)

The speed of the electrons can then be calculated using the formula:

v = √(2E/m)

Where: - v = speed of the electrons - E = maximum kinetic energy of the emitted electrons - m = mass of the electron (9.11 × 10⁻³¹ kg)

Let's proceed with the calculations.

First, we need to find the frequency of the incident light using the given wavelength:

c = fλ

Where: - c = speed of light in a vacuum (3 × 10⁸ m/s) - f = frequency of the light - λ = wavelength of the light

Calculations

1. Finding the Frequency of the Incident Light - Given: λ = 3 × 10⁻⁷ m - Using the formula: c = fλ - Solving for f: f = c/λ - Substituting the values: f = (3 × 10⁸ m/s) / (3 × 10⁻⁷ m) - Calculating: f ≈ 1 × 10¹⁵ Hz

2. Finding the Maximum Kinetic Energy of the Emitted Electrons - Given: Φ (work function of potassium) = 2.3 eV - Converting eV to Joules: 1 eV = 1.602 × 10⁻¹⁹ J - Φ = 2.3 eV = 2.3 × 1.602 × 10⁻¹⁹ J = 3.686 × 10⁻¹⁹ J - Using the formula: E = hf - Φ - Substituting the values: E = (6.626 × 10⁻³⁴ J·s) × (1 × 10¹⁵ Hz) - 3.686 × 10⁻¹⁹ J - Calculating: E ≈ 3.973 × 10⁻¹⁹ J

3. Finding the Speed of the Electrons - Using the formula: v = √(2E/m) - Substituting the values: v = √(2 × 3.973 × 10⁻¹⁹ J / 9.11 × 10⁻³¹ kg) - Calculating: v ≈ 6.57 × 10⁵ m/s - Converting to km/s: v ≈ 6.57 × 10² km/s

Answer

The speed of the electrons ejected from potassium by light with a wavelength of 3 × 10⁻⁷ m is approximately 6.57 × 10² km/s.