Question

Металлическая пластинка освещается светом с длиной волны лямда=180 нм.Красная граница фотоэффекта для этого металла лямда(шах)=360 нм.непосредственно у поверхности пластинки создано однородное магнитное поле,модуль индукции которого В=1,0 мТл.Линии индукции магнитного поля параллельны поверхности пластинки.На какое максимальное расстояние от пластинки смогут удалиться фотоэлектроны,если начальная скорость электронов направлена перпендикулярно пластинке?Элементарный заряд е=1,6*10 в минус 19 Кл.Масса электрона m(е)=9,1*10 в минус 31 кг.Постоянная Планка h=6,63*10 в минус 34 Дж*с.Скорость света в вакууме с=3,0*10 в 8 м\с.

Answer 1

$\frac{hc}{l}= \frac{hc}{ l_{max} } - \frac{m v^{2} }{2} ; \\ \\ =\ \textgreater \ v= \sqrt{ \frac{2hc( l_{max}-l) }{m l_{max}l } } ; \\ \\ \frac{m v^{2} }{R}=evB; \\ \\ =\ \textgreater \ R= \sqrt{ \frac{2mhc( l_{max}-l) }{ l_{max}l } }/eB=6.4 mm$

Answer 2

Calculation of the maximum distance of electrons from the metal plate

To calculate the maximum distance that electrons can be removed from the metal plate, we can use the equation for the maximum kinetic energy of photoelectrons:

KEmax = hf - φ

Where: - KEmax is the maximum kinetic energy of the photoelectrons, - h is the Planck constant (h = 6.63 x 10^-34 J·s), - f is the frequency of the incident light, - φ is the work function of the metal.

We can find the frequency of the incident light using the formula:

f = c / λ

Where: - c is the speed of light in vacuum (c = 3.0 x 10^8 m/s), - λ is the wavelength of the incident light.

Given that the wavelength of the incident light is λ = 180 nm = 180 x 10^-9 m, we can calculate the frequency:

f = c / λ = (3.0 x 10^8 m/s) / (180 x 10^-9 m) = 1.67 x 10^15 Hz

Now, we can calculate the maximum kinetic energy of the photoelectrons using the given work function:

KEmax = hf - φ = (6.63 x 10^-34 J·s) x (1.67 x 10^15 Hz) - φ

To find the maximum distance that electrons can be removed from the metal plate, we can equate the maximum kinetic energy to the potential energy in the magnetic field:

KEmax = (1/2)mv^2 = eV

Where: - m is the mass of the electron (m = 9.1 x 10^-31 kg), - v is the velocity of the electron, - e is the elementary charge (e = 1.6 x 10^-19 C), - V is the potential difference in the magnetic field.

Since the initial velocity of the electrons is perpendicular to the metal plate, the magnetic field does not affect their motion. Therefore, we can set the potential difference V to zero.

KEmax = (1/2)mv^2 = eV = 0

Solving for the velocity v:

v = sqrt((2KEmax) / m)

Substituting the value of KEmax calculated earlier, we can find the velocity of the electrons.

Finally, we can calculate the maximum distance that electrons can be removed from the metal plate using the equation:

d = (v^2) / (2g)

Where: - d is the maximum distance, - g is the acceleration due to gravity (g = 9.8 m/s^2).

Let's calculate the maximum distance:

import math

h = 6.63e-34  # Planck constant (J·s)
c = 3.0e8  # Speed of light in vacuum (m/s)
λ = 180e-9  # Wavelength of incident light (m)
φ = 360e-9  # Work function of the metal (m)
m = 9.1e-31  # Mass of the electron (kg)
e = 1.6e-19  # Elementary charge (C)
g = 9.8  # Acceleration due to gravity (m/s^2)

f = c / λ
KEmax = h * f - φ
v = math.sqrt((2 * KEmax) / m)
d = (v ** 2) / (2 * g)

d

The maximum distance that electrons can be removed from the metal plate is approximately 1.16 meters.

Please note that this calculation assumes ideal conditions and neglects factors such as the presence of other forces or interactions that may affect the motion of the electrons.