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course Phy 202
8/1 at 8:30
Problems 2,3 and 6 dont existProblem
A photoelectric metal has work function 1.04 electron volts. If it is exposed to monochromatic light with wavelength 410 nm, what is the magnitude of the maximum potential difference through which an ejected electron might travel?
Solution:
Frequency= 3x10^8/410x10^-9
=0.73x10^-15
energy= 6.63x10^-34 x 0.73x10^-15
= 4.9x10^-19
4.9x10^-19/1.6x10^-19
= 3.0
3.0-1.04=1.96 volts
The energy of a photon is h * f, where f is the frequency of the electromagnetic wave and h is Planck's constant 6.63 * 10^-34 J s.
The frequency of the wave is f = c / `lambda, where c = 3 * 10^8 m/s is the speed of light in a vacuum.
We therefore have
frequency of light: f = 3 * 10^8 m/s / 410 nm = 3 * 10^8 m/s / ( 410 * 10^-9 m) = .7317 * 10^15 Hz
and photon energy
photon energy = h * f = 6.63 * 10^-34 J s * .7317 * 10^15 Hz = 4.858 * 10^-19 Joules.
An electron volt is the energy gained by an electron as it travels through a potential difference of 0 volt; this energy is equal to the product of the charge 1.6 * 10^-19 C of an electron and the 1 volt = 1 J/C potential difference, or 1.6 * 10^-19 J:
1 eV = 1.6 * 10^-19 J.
Expressing this energy in units of electron volts we have
photon energy = 4.858 * 10^-19 J / (1.6 * 10^-19 J / eV) = 3.036.
When a conduction electron is ejected by a photon it can acquire the entire energy of the photon in the form of KE. However, to escape the metal surface requires a certain energy. This energy is called the work function, equal to 1.04 eV in the present example. It is possible for even more energy to be lost, and it is possible for a photon to give up less than its total energy to an electron, so not all electrons will escape the metal, and not all that escape will have the maximum possible KE as they leave the surface.
When a photon of wavelength 410 nm gives up its total 3.036 of energy to an electron, and when the electron loses no more than the required 1.04 eV, the electron will escape with KE
escape KE = 3.036 - 1.04 eV = 1.996 eV.
An electron with this KE can overcome a potential difference of
max PE overcome = 1.996 volts.
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Problem
A beam consisting of 60 eV electrons (electron mass 9.11 * 10^-31 kg) is incident on a thin wafer of a crystal with layer spacing 2.7 Angstroms. Surprisingly we find that the electrons, which are particles, are scattered in such a way as to form an interference pattern identical to that of a wave. What will be the distance between central interference maxima at a distance of 18 cm from the wafer?
Solution:
KE= 60 x 1.6x10^-19
= 0.96x10^-17
Momentum= sqrt(2x9.11x10^-31 x 0.96x10^-17)
= 4.2x10^-24
Wavelength= 6.63x10^-34/(4.2x10^-24
= 1.6x10^-10
Angle= sin^-1(1.6/2.7)
= 36.3 degrees
18 x sin(36.3)
= 10.65cm
A particle with momentum p behaves in many situations as a wave with wavelength
deBroglie wavelength: `lambda = h / p,
where h is Planck's constant. This wavelength is called the deBroglie wavelength after Louis deBroglie, who postulated this model in the 1920's.
To find the momentum of the electron from the given information we first find the KE in Joules:
The KE of the given electron in Joules is 60 eV * (1.6 * 10^-19 J / eV) = .96 * 10^-17 J.
We can find the momentum m v of the electron by multiplying its KE, which is .5 m v^2, by 2 m to obtain m^2 v^2, then taking the square root:
momentum of electron: p = m v = `sqrt( m^2 v^2) = `sqrt ( 2 * m * .5 mv^2) = `sqrt( 2 * m * KE), so
momentum = `sqrt[ (2 * 9.11 * 10^-31 kg) * ( .96 * 10^-17 J ) ] = 4.156922 * 10^-24 kg m/s.
The deBroglie wavelength of the electron is therefore
wavelength: `lambda = h / p = 6.63 * 10^-34 J s / ( 4.156922 * 10^-24 kg m/s ) = 1.59493 * 10^-10 m = 1.59493 * 10^-10 Angstroms.
The layers of the crystal effectively form slits through which the beam of electrons will diffract. Interference maxima will occur with path differences of 0, 1, 2, ... wavelengths. The central maximum will occur along the direction of the original beam, provided the crystal is oriented perpendicular to the beam. The next maximum will occur at an angle for which the path difference between paths from adjacent slits is one wavelength. If a stands for slit separation and `lambda for wavelength, this occurs at angle `theta such that
sin(`theta) = `lambda / a.
Thus the angle at which the maximum occurs is
`theta = sin^-1 ( 1.59493 Angstroms / 2.7 Angstroms) = 36.22612 degrees.
At a distance of 18 cm the separation of this angle from the central beam will be
separation of adjacent maxima = 18 cm * sin( 36.22612 degrees) = 13.17775 cm.
General Solution
An electron with kinetic energy E will have momentum m v = `sqrt( 2 m * (1/2 mv^2) ) = `sqrt( 2 m E ).
Its deBroglie wavelength will therefore be
deBroglie wavelength: `lambda = h / p = h / `sqrt( 2 m E ).
A beam of such electrons incident on a crystal with layer spacing a will exhibit interference maxima at angles `theta such that
sin(`theta) = n `lambda / a, n = 0, 1, 2, ... .
The first-order maximum (n=1) will thus occur at angle
`theta1 = sin^-1( `lambda / a) = sin^-1 ( h / [ a*( `sqrt( 2 m E) ].
We will observe at least the first-order maximum as long as h / [ a `sqrt( 2 m E) )]< 1. If E is too small, we will not observe a maximum.
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Set 57 Problem number 5
Problem
What is the approximate uncertainty in the velocity of an electron (mass 9.11 * 10^-31 kg) known to remain within a distance of 1.99 Angstroms of a proton?
What kinetic energy would the electron have at this velocity?
Approximately how many times would the electron 'orbit' the proton in a circular 'orbit' at this distance and velocity?
Compare the centripetal acceleration at this distance and velocity to the acceleration of the electron due to the electrostatic force between an electron and a proton at this distance.
Solution:
2 x 2= 4x10^-10 m
momentum uncertainty= 6.63x10^-34/4x10^-10
= 3.3x10^-24
velocity uncertainty= 3.3x10^-24/ 9.11x10^-34
= 0.37x10^-6
electron KE= 0.5(9.11x10^-31)(0.37x10^-6)^2
= 0.6x10^-19
Orbit time= 2 pi x 2x10^-10/ 0.37x10^-6
= 34x10^-16
Frequency= 1/34x10^-16
= 0.03x10^-16
Acceleration:
0.37x10^-6^2/ 2.0
=0.07x10^-22
Force= 9.11x10^-31 x 0.07x10^2-22
=0.6x10^-9
Magnitude=9x10^9 x 1.6x10^-19 / 2.0x10^-10
=5.8 x10^-9
The product of the uncertainty in the position (standard units of meters) and the uncertainty in momentum (standard units of kg m/s) is equal to Planck's constant 6.63 * 10^-34 J s (units kg m^2 / s^2 * s = kg m^2 / s, same as units of position multiplied by units of momentum). We write this as
uncertainty principle: `dx * `dp = h.
The position uncertainty 2 * 1.99 Angstroms = 3.98 * 10^-10 meters, the diameter of the circle on which the electron 'orbits'. This implies an uncertainty of
momentum uncertainty: `dp = h / `dx = ( 6.63 * 10^-34 kg m^s / s ) / ( 3.98* 10^-10 m) = 3.331 * 10^-24 kg m/s.
Since momentum is the product p = m v of mass and velocity, the uncertainty in velocity is
velocity uncertainty: `dv = `dp / m = 3.331 * 10^-24 kg m/s / (9.11 * 10^-34 ks) = .3656 * 10^6 m/s.
At this velocity the electron would have kinetic energy
electron KE: .5 m v^2 = .5 ( 9.11 * 10^-31 kn) * ( .3656 * 10^6 m/s)^2 = .6088 * 10^-19 J.
To find the time required to 'orbit' at the given distance, we will divide the circumference of the 'orbit' by the velocity:
time to orbit = circumference / velocity = 2 `pi * 1.99 * 10^-10 meters / ( .3656 * 10^6 m/s) = 34.16 * 10^-16 sec.
The frequency of the orbit is therefore
frequency = 1 / ( 34.16 * 10^-16 sec) = .02927 * 10^16 / sec.
The centripetal acceleration is
centripetal acceleration = v^2 / r = ( .3656 * 10^6 m/s) ^ 2 / ( 1.99 Angstroms) = ( .3656 * 10^6 m/s) ^ 2 / ( 1.99 * 10^-10 m ) = .06716 * 10^22 m/s^2.
The force on an electron with this acceleration is
centripetal force = m * centripetal acceleration = (9.11 * 10^-31 kg) * ( .06716 * 10^22 m/s^2) = .6118 * 10^-9 N.
The magnitude of the force of attraction between proton and electron is
| Coulomb force | = | k q1 q2 / r^2 | = 9 * 10^9 N m^2 / C^2 * (1.6 * 10^-19 C) ^ 2 / ( 1.99 * 10^-10 m)^2 = 5.818 * 10^-9 Newtons.
Note the order-or-magnitude agreement between the result from the uncertainty principle and the result from Coulomb's Law.
General Solution
A particle confined within some distance `dx = 2 * r will have uncertainty `dp in its momentum, where `dx * `dp = h (Planck's constant).
Thus we have `dp = h / `dx = h / (2 * r ).
If we know the mass of the particle we can find from the momentum `dp = m `dv the velocity corresponding to a momentum equal to the uncertainty:
`dv = `dp / m = h / ( 2 * r * m )
The KE of the particle will be
KE = .5 m v^2 = ( m v ) ^ 2 / ( 2 m ).
An electron moving at this velocity in a circular orbit at distance r from a proton would require time
time to orbit proton: `dt = circumference / velocity = 2 `pi r / ( h / 2 * r * m ) = 4 `pi r^2 mE / h
to complete an orbit, where mE is the mass of an electron..
The centripetal force required to hold the electron in the orbit is
centripetal force = m v^2 / r = mE ( ( 4 `pi r^2 mE / h )^2 ) / r = 16 `pi^2 mE^3 r^3 / h^2.
This can be compared to the Coulomb attraction between electron and proton:
| Coulomb force | = k qE^2 / r^2,
where qE is the magnitude of the fundamental charge.
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