Combustion is an example of ______ to______

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 kinetic energy.

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 minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and its value depends on the metal. 

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 frequency has an inverse relation with the wavelength

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (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

Use the concept of fields to describe how forces can act from a distance.

Leni [432]

The strength of a field force changes with distance from the sourceof the field-stronger closer to the source, weaker farther away from the source. The source can be either a mass, a charged particle, or a magnetic pole.

8 0

3 years ago

How does light reflect off of mirrors and other types of surfaces?

Marat540 [252]

Light can reflect from mirrors because mirrors are a prism.

4 0

3 years ago

Read 2 more answers

A droplet of ink in an industrial ink-jet printer carries a charge of 1.6 x 10–11 coulombs and is deflected onto paper by a forc

Novosadov [1.4K]

Answer:

The Strength of the electric field produced = 2 × 10⁷ N/C

Explanation:

Electric Field: This is defined as the region where an electric force is experienced.

Electric Field Strength: The intensity of an electric field at any point is defined as the force per unit charge which it exert at that point. It direction is that of the force exerted on a positive charge.

It is represented mathematically as,

E = F/Q ................................. Equation 1

Where E = Electric field strength, F = electric force, Q = test charge.

Given: F = 3.2 × 10⁻⁴ N, Q = 1.6 × 10⁻¹¹ C

Substituting these values into equation 1

E= 3.2 × 10⁻⁴/1.6 × 10⁻¹¹

E= 2 × 10⁷ N/C

Thus the Strength of the electric field produced = 2 × 10⁷ N/C

3 0

3 years ago

Thus, an object with a momentum of 0 (does or does not have

creativ13 [48]

Answer:

At momentum zero, kinetic energy = zero

At momentum 10, kinetic energy can never be equal to zero.

Explanation:

An object is said to have zero momentum when either its mass or its velocity is zero. Since, negetive mass doesn't exist, the velocity should be zero and this case, the object must be at rest. Momentum is zero when change in velocity is zero, but kinetic can never be zero when body is motion or has a momentum of 10.

5 0

3 years ago

Combustion is an example of ______ to_______ energy conversion.

Combustion is an example of to_ energy conversion.