Question

Красная черта фотоэффекта в цезия равна 654 нм. найдите скорость вылета фотоэлектронов при облучении цезия лучами с длиной волны 500 нм

Answer 1

E(фотона) = A+E(кинетическая)

E(ф)-A=E(k)

 

E(k)=mv^2/2

E(ф)=hc/(на длину волны)

A=hс/(длину волны (кр.) )

 

h=6,62*10^-34       c=3*10^8              m= 9.1*10^-31

 

Конечная формула

 

Answer 2

Calculating the Speed of Photoelectrons Emitted from Cesium

To calculate the speed of the photoelectrons emitted from cesium when irradiated with light of a wavelength of 500 nm, we can use the formula for the kinetic energy of photoelectrons:

Kinetic Energy of Photoelectrons (E) = Planck's constant (h) × frequency of the incident light (f) - Work Function (φ)

The kinetic energy of the photoelectrons can be equated to their kinetic energy:

E = 1/2 × m × v^2

Where: - m is the mass of the photoelectron - v is the speed of the photoelectron

By equating these two expressions for the kinetic energy, we can solve for the speed of the photoelectron.

Calculation

First, we need to find the frequency of the incident light using the given wavelength of 500 nm.

The speed of light (c) = 3.00 × 10^8 m/s

The frequency (f) can be calculated using the formula:

f = c / λ

Where: - c is the speed of light - λ is the wavelength of the light

Substituting the given values: f = (3.00 × 10^8 m/s) / (500 × 10^-9 m)

Now, we can calculate the kinetic energy of the photoelectrons using the frequency of the incident light and the work function of cesium.

Conclusion

By using the calculated kinetic energy, we can then solve for the speed of the photoelectrons emitted from cesium when irradiated with light of a wavelength of 500 nm.