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3 years ago

12

A mass spectrometer applies a voltage of 2.00 kV to accelerate a singly charged positive ion. A magnetic field of B = 0.400 T th

en bends the ion into a circular path of radius 0.305 m. What is the mass of the ion?

1 answer:

N76 [4]3 years ago

3 0

The mass of the ion is 5.96 X 10⁻²⁵ kg

<u>Explanation:</u>

The electrical energy given to the ion Vq will be changed into kinetic energy $\frac{1}{2}mv^2$

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force $\frac{mv^2}{r}$ .

So,

$Vq = \frac{1}{2}mv^2$

and

$Bqv = \frac{mv^2}{r}$

Right from these eliminating v, we can derive

$m = \frac{B^2r^2q}{2V}$

On substituting the value, we get:

$m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\$

m = 5.96 X 10⁻²⁵ kg.

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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

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Solution :

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Also, force of friction is given by :

$f = \mu mg$ ....2)

Equating equation 1) and 2), we get :

$\mu mg = ma\\\\\mu = \dfrac{a}{g}\\\\\mu = \dfrac{2.42}{9.8}\\\\\mu = 0.247$

Therefore, the coefficient of kinetic friction is 0.247 .

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To solve this problem it is necessary to apply the concepts related to Slit Diffraction.

The expression for separation between fringes is given by,

$d= \frac{\lambda D} {a}$

Where,

$\lambda =$ Wavelength

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a = Slit width

D = Distance between the slits

Re-arrange to find $\lambda$ , we have that

$\lambda = \frac{da}{D}$

Replacing with our values we have

$\lambda = \frac{(0.005575)(4*10^-5)}{0.3}$

$\lambda = 743.15*10^{-9}$

$\lambda = 743.14nm$

Therefore the wavelength of the laser you used to collect the data is 743.14nm

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3 years ago