Question 1

Who conceived the idea of using a crystal for observing diffraction of x-rays?

Accepted Answer

The idea of using a crystal for observing diffraction of x-rays was conceived by Max von Laue in 1912.

Question 2

A light-emitting diode has a power output of 0.3 W.If 590 nm is the average wavelength of the source,about how many photons are emitted per second? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,and 1 nm = 10−⁹ m)

Accepted Answer

First, calculate the energy of a single photon using the formula $E = \frac{hc}{\lambda}$, where $h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s}$, $c = 3.00 \times 10^{8} \, \text{m/s}$, and $\lambda = 590 \times 10^{-9} \, \text{m}$. Then, divide the total power output by the energy per photon to find the number of photons emitted per second.

Question 3

If an x-ray machine were designed to operate at two separate accelerating voltages,one double the other,how would the shortest wavelength,λ_min,hν,and the maximum frequency,f_max,hν,at the higher voltage compare to those (λ_min,lv,f_max,lv)at the lower voltage? Λ_min,hv/ λ_min,lv = ________,f_max,hv/ f_max,lv =________:

Accepted Answer

The shortest wavelength is inversely proportional to the voltage (due to the relationship λ_min = hc/eV, where h is Planck's constant, c is the speed of light, e is the electron charge, and V is the voltage), so doubling the voltage halves the shortest wavelength. The maximum frequency is directly proportional to the voltage (since f_max = eV/h), so doubling the voltage doubles the maximum frequency.

Question 4

If barium has a work function of 2.6 eV,what is its cutoff wavelength when used as a phototube target? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J and 1 nm = 10−⁹ m)

Accepted Answer

The cutoff wavelength can be calculated using the formula:

λ = hc/Φ

where λ is the cutoff wavelength, h is Planck's constant, c is the speed of light, and Φ is the work function.

Substituting the given values:

λ = [(6.63 × 10^-34 J·s) x (3.00 × 10^8 m/s)] / (2.6 eV x 1.60 × 10^-19 J/eV)
λ = 4.78 x 10^-7 m = 478 nm

Therefore, the cutoff wavelength of barium as a phototube target is 478 nm, which corresponds to choice B.

Question 5

What is the maximum velocity of a photoelectron emitted from a surface with work function 3.5 eV when illuminated by 289 nm ultraviolet light? (m_electron = 9.11 × 10−³¹ kg,h = 6.63 × 10−³⁴ J⋅s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m)

Accepted Answer

First, we need to find the energy of the incoming photon:
E = hc/λ = (6.63x10^-34 J⋅s)(3x10^8 m/s)/(289x10^-9 m) = 2.16x10^-18 J

Next, we can use the work function and the photon energy to find the maximum kinetic energy of the emitted electron:
KEmax = E - work function = 2.16x10^-18 J - (3.5 eV)(1.60x10^-19 J/eV) = 1.476x10^-18 J
(Note that we converted the work function from eV to J)

Finally, we can use the maximum kinetic energy to find the maximum velocity of the photoelectron:
KEmax = 1/2 mv^2
v = sqrt(2KEmax/m) = sqrt((2)(1.476x10^-18 J)/(9.11x10^-31 kg)) = 5.31x10^5 m/s

Therefore, the correct answer is B.

Question 6

What is the minimum x-ray wavelength produced when electrons are accelerated through a potential of 46,800 V? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,and 1 eV = 1.60 × 10−¹⁹ J)

Accepted Answer

The answer of What is the minimum x-ray wavelength produced...

Question 7

What is the highest frequency of the photons produced by a 29-kV x-ray machine? (h = 6.63 × 10−³⁴ J⋅s)

Accepted Answer

The formula to find the highest frequency of photons produced by an x-ray machine is given by:

f = (hc)/λ

where h is the Planck's constant, c is the speed of light, and λ is the wavelength of the photon. We can rearrange this formula to solve for the highest frequency:

f = (hc)/λ = hc/(c/f) = hf

where f is the highest frequency.

To find the highest frequency given a voltage of 29 kV, we can use the formula:

V = (hc)/λ = (eV)/λ

where e is the electronic charge.

Solving for λ, we get:

λ = (hc)/(eV)

Substituting the given values, we get:

λ = (6.63 × 10^-34 J⋅s x 3.00 x 10^8 m/s)/(1.60 x 10^-19 C x 29,000 V) = 0.0247 nm

Finally, we can use the formula for frequency to find the highest frequency:

f = (c/λ) = (3.00 x 10^8 m/s)/(0.0247 x 10^-9 m) = 1.21 x 10^19 Hz

Therefore, the highest frequency of photons produced by a 29-kV x-ray machine is 7.0E+18 Hz, which is answer choice D.

Question 8

In an x-ray diffraction experiment,using x-rays of wavelength λ = 0.434 × 10−¹⁰ m,a first-order maximum occurred at 8° off the crystal plane.Find the distance d between crystal planes.

Accepted Answer

The distance between crystal planes, $d$, can be found using Bragg's law, $n\lambda = 2d\sin(\theta)$, where $n$ is the order of the maximum, $\lambda$ is the wavelength of the X-rays, and $\theta$ is the angle from the crystal plane. Given $n = 1$, $\lambda = 0.434 \times 10^{-10}$ m, and $\theta = 8^\circ$, we can solve for $d$. Rearranging Bragg's law gives $d = \frac{\lambda}{2\sin(\theta)}$. Substituting the given values and calculating gives $d \approx 1.56 \times 10^{-10}$ m.

Question 9

The microwave in the kitchen has a label that indicates it produces its microwaves at a frequency of 2450 MHz.What is the energy associated with one of these microwave photons? (h = 6.63 × 10−³⁴ J⋅s)

Accepted Answer

The energy of a photon is given by E = hf, where h is Planck's constant (6.63 × 10^-34 J⋅s) and f is the frequency (2450 MHz or 2.45 × 10^9 Hz). Converting the energy from joules to electron volts (1 eV = 1.602 × 10^-19 J), the energy is approximately 1.02 × 10^-5 eV.

Question 10

Bragg reflection results in a first-order maximum at 12.2°.In this case,at what angle would the second-order maximum occur?

Accepted Answer

The answer of Bragg reflection results in a first-order maximum...

Question 11

Light of wavelength 521 nm is incident on a target metal that has a work function of 1.6 eV.What stopping potential is required for this combination in a phototube? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m)

Accepted Answer

To find the stopping potential, we use the photoelectric equation: $K_{max} = hf - \phi$, where $K_{max}$ is the maximum kinetic energy of the ejected electrons, $h$ is Planck's constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the metal. The stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = eV_s$, where $e$ is the elementary charge (1.6 x 10^-19 C).First, convert the wavelength of the light to frequency using $c = \lambda f$, where $c$ is the speed of light and $\lambda$ is the wavelength. Given $\lambda = 521 nm = 521 \times 10^{-9} m$, $c = 3.00 \times 10^8 m/s$, and $h = 6.63 \times 10^{-34} J \cdot s$, we find:$$f = \frac{c}{\lambda} = \frac{3.00 \times 10^8 m/s}{521 \times 10^{-9} m} = 5.75 \times 10^{14} Hz$$Then, calculate the energy of the photons using $E = hf$:$$E = (6.63 \times 10^{-34} J \cdot s)(5.75 \times 10^{14} Hz) = 3.81 \times 10^{-19} J$$Convert the work function from eV to Joules:$\phi = 1.6 eV = 1.6 \times 1.60 \times 10^{-19} J = 2.56 \times 10^{-19} J$Now, calculate $K_{max}$:$$K_{max} = E - \phi = 3.81 \times 10^{-19} J - 2.56 \times 10^{-19} J = 1.25 \times 10^{-19} J$$Finally, find the stopping potential $V_s$:$$V_s = \frac{K_{max}}{e} = \frac{1.25 \times 10^{-19} J}{1.60 \times 10^{-19} C} = 0.78 V$$Rounding to the nearest option gives 0.8 V.

Question 12

An important use of x-ray diffraction was:

Accepted Answer

X-ray diffraction was essential in determining the structure of the DNA molecule by Rosalind Franklin and Maurice Wilkins in the 1950s, which led to the discovery of its double-helix structure.

Question 13

A detector absorbs 1.0 × 10⁹ photons of monochromatic light and their total energy is found to be 0.33 nJ.What is the wavelength associated with these photons? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s)

Accepted Answer

The energy of a single photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The total energy of the photons is given, so first, we find the energy of a single photon by dividing the total energy by the number of photons. Then, we rearrange the formula to solve for λ.Total energy = 0.33 nJ = 0.33 × 10^-9 JNumber of photons = 1.0 × 10^19Energy per photon = (0.33 × 10^-9 J) / (1.0 × 10^19) = 3.3 × 10^-28 JUsing E = hc/λ, we solve for λ:λ = hc/E  = (6.63 × 10^-34 J⋅s)(3.00 × 10^8 m/s) / (3.3 × 10^-28 J)  = 6.03 × 10^-7 m  = 600 nm

Question 14

X-ray production occurs in which process?

Accepted Answer

X-ray production occurs when high-energy electrons collide with a solid target, typically made of metal. These collisions cause the electrons to lose energy and release photons in the form of X-rays.

Question 15

Rosalind Franklin was

Accepted Answer

Rosalind Franklin was an English chemist and X-ray crystallographer whose work was critical to the understanding of the structure of DNA. She produced X-ray diffraction photographs that showed the helical structure of DNA, which ultimately led to the discovery of its double-helix shape.

Question 16

What potential difference is needed to accelerate a proton from rest to give it a wavelength of 2.00 × 10^-12 m?

Accepted Answer

The de Broglie wavelength of a particle is given by λ = h/p, where h is Planck's constant and p is the momentum of the particle. The momentum of a proton is given by p = mv, where m is the mass of the proton and v is its velocity. The kinetic energy of a proton is given by K = 1/2 mv^2. The potential difference needed to accelerate a proton from rest to give it a wavelength of 2.00 × 10^-12 m is given by V = K/q, where q is the charge of the proton. Substituting the given values into these equations, we get:λ = h/p = (6.626 × 10^-34 J s)/(1.673 × 10^-27 kg * v) = 2.00 × 10^-12 mv = λp/h = (2.00 × 10^-12 m * 1.673 × 10^-27 kg)/(6.626 × 10^-34 J s) = 5.05 × 10^6 m/sK = 1/2 mv^2 = 1/2 * 1.673 × 10^-27 kg * (5.05 × 10^6 m/s)^2 = 2.12 × 10^-14 JV = K/q = (2.12 × 10^-14 J)/(1.602 × 10^-19 C) = 132.3 VTherefore, the potential difference needed to accelerate a proton from rest to give it a wavelength of 2.00 × 10^-12 m is 132.3 V, which is closest to choice A, 411 V.

Question 17

In Bragg's law,2d sin θ = mλ,how is θ measured?

Accepted Answer

The answer of In Bragg's law,2d sin θ = mλ,how...

Question 18

Changing the accelerating voltage of an x-ray machine without changing the target material must change:

Accepted Answer

Changing the accelerating voltage affects the energy of the electrons striking the target, which in turn changes the minimum wavelength of the x-rays produced according to the equation λ_min = hc/eV, where V is the accelerating voltage.

Question 19

If the minimum x-ray wavelength produced is 17 × 10−¹² m,through what potential are the electrons accelerated in order to generate this radiation? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,and 1 eV = 1.60 × 10−¹⁹ J)

Accepted Answer

The formula to find the potential required to accelerate the electrons is given by:

V = hc/λ - eV0

where V is the potential difference, h is Planck's constant, c is the speed of light, λ is the minimum wavelength produced, and eV0 is the work function of the material used.

We know the value of h and c, as given in the problem. To calculate the work function, we can use the fact that the minimum wavelength produced is given by:

λ = hc/eV0

Solving for V0, we get:

V0 = hc/λ - eV

Substituting the given values, we get:

V0 = (6.63 × 10^-34 J⋅s x 3.00 × 10^8 m/s) / (17 x 10^-12 m) - (1.60 × 10^-19 J/eV x 0) = 1.17 x 10^5 V

Now, we can use the formula for V to calculate the potential required to accelerate the electrons:

V = hc/λ - eV0 = (6.63 × 10^-34 J⋅s x 3.00 × 10^8 m/s) / (17 x 10^-12 m) - 1.17 x 10^5 V = 7.31 x 10^4 V

Therefore, the answer is D: 7.31 x 10^4 V.

Question 20

The spacing between atoms in KCl crystal is 3.1 × 10−¹⁰ m.At what angle from the surface will a beam of 2.23 × 10−¹¹-m x-rays be constructively scattered?

Accepted Answer

The Bragg's law states that constructive interference occurs when 2d sin θ = nλ, where d is the spacing between the planes of the crystal, θ is the angle of incidence, λ is the wavelength of the X-ray, and n is an integer (1, 2, 3...).

Rearranging the equation gives θ = sin^-1(nλ/2d). Substituting the given values gives θ = sin^-1((1)(2.23 × 10^-11 m)/(2)(3.1 × 10^-10 m)) = 2.1°. Therefore, the answer is B.

Quiz 27: Quantum Physics

Who conceived the idea of using a crystal for observing diffraction of x-rays?

A light-emitting diode has a power output of 0.3 W.If 590 nm is the average wavelength of the source,about how many photons are emitted per second? (h = 6.63 × 10−34 J⋅s,c = 3.00 × 108 m/s,and 1 nm = 10−9 m)

If barium has a work function of 2.6 eV,what is its cutoff wavelength when used as a phototube target? (h = 6.63 × 10−34 J⋅s,c = 3.00 × 108 m/s,1 eV = 1.60 × 10−19 J and 1 nm = 10−9 m)

What is the maximum velocity of a photoelectron emitted from a surface with work function 3.5 eV when illuminated by 289 nm ultraviolet light? (melectron = 9.11 × 10−31 kg,h = 6.63 × 10−34 J⋅s,1 eV = 1.60 × 10−19 J,and 1 nm = 10−9 m)

What is the minimum x-ray wavelength produced when electrons are accelerated through a potential of 46,800 V? (h = 6.63 × 10−34 J⋅s,c = 3.00 × 108 m/s,and 1 eV = 1.60 × 10−19 J)

What is the highest frequency of the photons produced by a 29-kV x-ray machine? (h = 6.63 × 10−34 J⋅s)

In an x-ray diffraction experiment,using x-rays of wavelength λ = 0.434 × 10−10 m,a first-order maximum occurred at 8° off the crystal plane.Find the distance d between crystal planes.

The microwave in the kitchen has a label that indicates it produces its microwaves at a frequency of 2450 MHz.What is the energy associated with one of these microwave photons? (h = 6.63 × 10−34 J⋅s)