Question

Работа выхода электронов на поверхность бромистого серебра равна 4 эв. Определить красный предел фотоэффекта.​

Answer 1

Объяснение:

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Answer 2

Red Limit of the Photoelectric Effect

The red limit of the photoelectric effect can be determined based on the work function of a material. The work function is the minimum energy required to remove an electron from the surface of a material. In this case, we are given that the work function for bromide silver is 4 eV.

The red limit refers to the minimum frequency or maximum wavelength of light that can cause the photoelectric effect. When light with a frequency or wavelength below the red limit is incident on a material, it does not have enough energy to overcome the work function and eject electrons.

To determine the red limit, we can use the equation:

E = hf

where: - E is the energy of a photon - h is Planck's constant (approximately 6.626 x 10^-34 J·s) - f is the frequency of the light

Since we are given the work function in electron volts (eV), we need to convert it to joules (J) by multiplying by the elementary charge (e) of an electron, which is approximately 1.602 x 10^-19 C.

4 eV * (1.602 x 10^-19 C/e) = 6.408 x 10^-19 J

Now, we can rearrange the equation to solve for the frequency:

f = E / h

f = (6.408 x 10^-19 J) / (6.626 x 10^-34 J·s)

Calculating this, we find:

f ≈ 9.67 x 10^14 Hz

Therefore, the red limit of the photoelectric effect for bromide silver is approximately 9.67 x 10^14 Hz.

Please note that the sources provided did not directly answer the question, but the calculation is based on the principles of the photoelectric effect and the given work function of bromide silver.