Question

При уменьшении в 2 раза длины волны света, падающего на металлическую пластинку, максимальная кинетическая энергия электронов увеличилась в 3 раза. Определите работу выхода электронов, если первоначальная энергия фотонов равнялась 10 эВ.

Answer 1

По з-ну фотоэффекта:
Аф=Ав + Ек
Тогда, если длина света падающего уменьшается в 2 раза, а Eк увеличивается в 3 раза, следует:
Ав=Аф/2=5 эВ

Ответ: 5эВ

Answer 2

Calculating the Work Function and Kinetic Energy of Electrons

When the wavelength of light incident on a metallic plate is halved, and the maximum kinetic energy of the electrons is tripled, we can calculate the work function of the metal using the given initial energy of the photons.

Given: - Initial energy of photons (E) = 10 eV - Increase in maximum kinetic energy of electrons = 3 times - Halving of the wavelength of light

We can use the following formula to calculate the work function (Φ) and the maximum kinetic energy of the electrons (Kmax):

Work Function (Φ) = E - Kmax

Where: - E is the initial energy of the photons - Kmax is the maximum kinetic energy of the electrons

First, we need to calculate the new energy of the photons after halving the wavelength, and then we can calculate the work function and the maximum kinetic energy of the electrons.

Calculating New Energy of Photons

The energy of a photon can be calculated using the formula: E = hc / λ Where: - h is the Planck constant (6.626 x 10^-34 J·s) - c is the speed of light in a vacuum (3.00 x 10^8 m/s) - λ is the wavelength of the light

Given that the initial energy of the photons (E) is 10 eV, we can convert this to joules and then use the above formula to find the new energy of the photons after halving the wavelength.

Calculating Work Function and Maximum Kinetic Energy of Electrons

Once we have the new energy of the photons, we can use it to calculate the work function and the maximum kinetic energy of the electrons using the formula mentioned earlier.

Let's proceed with the calculations.