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Alex Ar [27]
3 years ago
13

(An easy problem which will be graded). Later in the quarter we will spend some time solving the diffusion equation Op(r, t) = D

V4p(r, t), at where D is called the diffusion constant. For now, let us consider a solution which represents a steady state solution (i.e apat = 0) and which has spherical symmetry. In that case the above reduces to 1d (20p) = 0 p2 dr dr / a) Obtain the solution of this second-order differential equation. It will have two arbitrary constants. b) Application: diffusion from a large distance to a spherical absorber. Consider p(r) to represent the density of some particles which are diffusing through some media. Think of them as food for a cell. Their density at large distance (infinity) is po. There is a spherical absorber of radius R centered at the origin, which means p(R) = 0. (This absorber represents the cell.) Show that the particular solution that satisfies these two condition is p(r) = po(1 – R/r). An interesting result: The current density (the number of particles passing through unit area per unit time) is given by j(r) = -Ddp/dr = -DpoR/r2. (The minus sign means that the flow is inward.) The magnitude of the current at a distance r is I = 47r2|j(r)= 47 DPOR. Note that this result says that the rate at which particles are absorbed is proportional to the radius of the sphere, not to the area of the sphere as would be the case if we were dealing with light being absorbed. Of course light does not travel by diffusion!

Physics
1 answer:
alexandr402 [8]3 years ago
7 0

Answer:

Explanation:

Answer is in the attachment below:

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Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

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E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

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Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

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So, the metal is Lithium.

B. cesium, potassium, sodium

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\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

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E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

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Cesium

Potassium

Sodium

C. 4.9 eV

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- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

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So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

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brilliants [131]

Answer:

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