Question

Квант длиной волны L = 342 Å(ангстрем) вырывает с чистой поверхности металлического лития фотоэлектрон, который описывает в магнитном поле напряженностью Н = 1,2*10^3 А/м окружность радиусом R = 1,2 см. Определить энергию (работу выхода), затраченную на освобождение данного электрона из атома лития.

Answer 1

Некоторые формулы по вашему мнению могут не подходить, я их просто вспоминал и писал, в итоге вышел на финишную прямую.
Осталось все подставить и решить, общая формула есть и все подписи )
удачи )

Answer 2

Calculation of the Energy (Work Function) Required to Release an Electron from a Lithium Atom

To determine the energy (work function) required to release an electron from a lithium atom, we need to consider the given parameters: the wavelength of the incident light (L = 342 Å) and the magnetic field strength (H = 1.2 * 10^3 A/m) in which the electron is moving in a circular path with a radius (R = 1.2 cm).

The energy required to release an electron from an atom is given by the equation:

Energy = Planck's constant * Speed of light / Wavelength

Let's calculate the energy required step by step:

1. Convert the wavelength from angstroms (Å) to meters (m): - 1 Å = 1 * 10^-10 m - L = 342 Å = 342 * 10^-10 m

2. Substitute the values into the equation: - Energy = (Planck's constant * Speed of light) / Wavelength - Energy = (6.626 * 10^-34 J*s * 3 * 10^8 m/s) / (342 * 10^-10 m)

3. Simplify the equation: - Energy = (6.626 * 3 * 10^-34 * 10^8) / 342 J - Energy = (19.878 * 10^-26) / 342 J - Energy = 5.8 * 10^-29 J

Therefore, the energy (work function) required to release the electron from the lithium atom is approximately 5.8 * 10^-29 J.

Please note that the sources provided do not directly address the specific question. However, the calculation is based on fundamental principles of quantum mechanics and the equation for energy of a photon.