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poizon [28]
3 years ago
11

A vector has what and direction

Physics
2 answers:
galben [10]3 years ago
8 0
A vector Quantity has MAGNITUDE and direction, so the ‘what’ is Magnitude.
lys-0071 [83]3 years ago
4 0
Vector, in physics, a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude. Although a vector has magnitude and direction, it does not have position.
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3 years ago
A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr
slega [8]
<h2>Answer: 1.011 eV</h2>

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.  

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a <u>kinetic energy. </u>

This is what Einstein proposed:  

Light behaves like a stream of particles called photons with an energy  E:

E=h.f (1)  

So, the energy E of the incident photon must be equal to the sum of the Work function \Phi of the metal and the kinetic energy K of the photoelectron:  

E=\Phi+K (2)  

Where \Phi is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal.  </u>

In this case \Phi=2eV  and K_{1}=4eV

So, for the first light source of wavelength \lambda_{1}, and  applying equation (2) we have:

E_{1}=2eV+4eV   (3)  

E_{1}=6eV   (4)  

Now, substituting (1) in (4):  

h.f=6eV (5)  

Where:  

h=4.136(10)^{-15}eV.s is the Planck constant

f is the frequency  

Now, the <u>frequency has an inverse relation with the wavelength </u>

\lambda_{1}:  

f=\frac{c}{\lambda_{1}} (6)  

Where c=3(10)^{8}m/s is the speed of light in vacuum  

Substituting (6) in (5):  

\frac{hc}{\lambda_{1}}=6eV (7)  

Then finding \lambda_{1}:  

\lambda_{1}=\frac{hc}{6eV } (8)  

\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}  

We obtain the wavelength of the first light suorce \lambda_{1}:  

\lambda_{1}=2.06(10)^{-7}m   (9)

Now, we are told the second light source \lambda_{2}  has the double the wavelength of the first:

\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)   (10)

Then: \lambda_{2}=4.12(10)^{-7}m   (11)

Knowing this value we can find E_{2}:

E_{2}=\frac{hc}{\lambda_{2}}   (12)

E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}   (12)

E_{2}=3.011eV   (13)

Knowing the value of E_{2} and \lambda_{2}, and knowing we are working with the same work function, we can finally find the maximum kinetic energy K_{2} for this wavelength:

E_{2}=\Phi+K_{2} (14)  

K_{2}=E_{2}-\Phi (15)  

K_{2}=3.011eV-2eV  

K_{2}=1.011 eV  This is the maximum kinetic energy for the second light source

7 0
3 years ago
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