Question

Работа выхода электронов из золота равна 4,76 эВ. Определите красную границу фотоэффекта для золота.​

Answer 1

В красной границе фотоэффекта:

$hv_{min}=A$

Частота связана с длиной волны формулой:

$v = \frac{c}{I}$

где c – скорость света. Таким образом:

$I = \frac{hc}{A}$

Вычислим, считая постоянную Планка равной h=4,13·10-15 эВ·с:

$I = \frac{4.13 \times 10 {}^{ - 15 } \times 3 \times 10 {}^{8} }{4.76} = 2.6 \times 10 {}^{ - 7} = 260$

Ответ: 260 нм

Answer 2

Red Edge of the Photoelectric Effect for Gold

The work function of gold, which is the minimum energy required to remove an electron from the surface of gold, is given as 4.76 eV. To determine the red edge of the photoelectric effect for gold, we need to find the wavelength of light that corresponds to this energy.

The relationship between energy and wavelength is given by the equation:

E = hc/λ

Where: - E is the energy of a photon - h is Planck's constant (6.626 x 10^-34 J·s) - c is the speed of light (2.998 x 10^8 m/s) - λ is the wavelength of light

To find the wavelength corresponding to the energy of 4.76 eV, we can convert the energy to joules and substitute the values into the equation:

4.76 eV = (4.76 x 1.6 x 10^-19 J) = hc/λ

Simplifying the equation, we can solve for λ:

λ = hc/E

Substituting the values:

λ = (6.626 x 10^-34 J·s x 2.998 x 10^8 m/s) / (4.76 x 1.6 x 10^-19 J)

Calculating this expression gives us the wavelength of light corresponding to the red edge of the photoelectric effect for gold.

Please note that the search results did not provide a direct answer to the specific wavelength for the red edge of the photoelectric effect for gold. However, the calculation provided above should give you the approximate value.