Question

Красная граница фотоэффекта для лития λкр = 540 нм. Максимальная скорость вылета электронов равна 10^6 м/с. Какова частота света, которым освещается катод? ПОДРОБНО ПОЖАЛУЙСТА

Answer 1

Запишем основное уравнение Эйнштейна для фотоэффекта:


hv= A+(mu^2/2)


v= (A+(mu^2/2))/h


v= ((ch/λ)+((mu^2)/2))/h


v= 12*10^(-14) Гц   .


Ответ:

v= 12*10^(-14) Гц   .

Answer 2

Calculation of the Frequency of Light

To determine the frequency of light illuminating the cathode, we can use the equation that relates the speed of light, wavelength, and frequency:

c = λ * f

Where: - c is the speed of light (approximately 3 x 10^8 m/s) - λ is the wavelength of light (given as 540 nm) - f is the frequency of light (what we need to find)

To find the frequency, we rearrange the equation:

f = c / λ

Now we can substitute the given values into the equation:

f = (3 x 10^8 m/s) / (540 x 10^-9 m)

Simplifying the expression:

f = (3 x 10^8) / (540 x 10^-9) Hz

Calculating the value:

f ≈ 5.56 x 10^14 Hz

Therefore, the frequency of the light illuminating the cathode is approximately 5.56 x 10^14 Hz.