Introduction To Health Physics
Introduction To Health Physics
5th Edition
ISBN: 9780071835275
Author: Johnson, Thomas E. (thomas Edward), Cember, Herman.
Publisher: Mcgraw-hill Education,
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Chapter 2, Problem 2.15P

(a)

To determine

The voltage required to accelerate the proton.

(a)

Expert Solution
Check Mark

Answer to Problem 2.15P

The voltage required to accelerate the proton from zero velocity to a velocity corresponding to a de Broglie wavelength of 0.01A is 816eV .

Explanation of Solution

Given:

De-Broglie Wavelength, λ=0.01A=1×1012m

Mass of the proton, mp=1.67×1027kg

Initial velocity, vi=0m/s

Formula used:

The de-Broglie wavelength of the particle is given as

  λ=hmv

Here,

  h=6.64×1034J-s is Planck’s constant

  m is a mass of the particle

  v is the velocity of the particle

Calculation:

Calculating the velocity of the proton corresponding to a wavelength of 0.01A , considering the relativistic effect

  λ=hmv

Plugging the values in the above equation:

  1×1012=6.64× 10 34mvmv=6.63×1022kg-m/s

We need to check whether the relativistic effect is important

  p=mv=mo 1 β 2 βc6.63×1022=1.627× 10 27×β×3× 108 1 β 2 β=vc=1.32×103<0.1v=3.95×105m/s

The voltage required to accelerate the proton from zero velocity to a velocity corresponding to a de Broglie wavelength of 0.01A is calculated as

  ΔPE=ΔKEqV=12m(vf2vi2)1.6×1019×V=12×1.67×1027( ( 3.95× 10 5 )202)V=816eV

Conclusion:

Thus,the voltage required to accelerate the proton from zero velocity to a velocity corresponding to a de Broglie wavelength of 0.01A is 816eV .

(b)

To determine

The kinetic energy of an electron corresponding to a de Broglie wavelength of 0.01A .

(b)

Expert Solution
Check Mark

Answer to Problem 2.15P

The kinetic energy of an electron corresponding to a de Broglie wavelength of 0.01A is 0.83MeV .

Explanation of Solution

Given:

De-Broglie Wavelength, λ=0.01A=1×1012m

Mass of the electron, me=9.1×1031kg

Formula used:

Kinetic energy is given as

  KE=12mv2

Here,

  m is a mass of the particle

  v is the velocity of the particle

Calculation:

The velocity of the electron corresponding to a wavelength of 0.01A

  λ=hmv

Plugging the values in the above equation

  1×1012=6.64× 10 349.1× 10 31×vv=7.296×108m/s>3×108

Since the speed of the electron is greater than the speed of light in a vacuum, therefore, we need to consider the effect of relativity

We need to check whether the relativistic effect is important

  p=mv=mo 1 β 2 βc6.63×1022=9.11× 10 31×β×3× 108 1 β 2 β=vc=0.92>0.1v=2.76×108m/s

The kinetic energy of the electron corresponding to a wavelength of 0.01A is calculated as:

  ΔKE=moc2(1 1 β 2 1)=9.1×1031×(3× 10 8)2(1 1 0.92 2 1)=1.27×1013J=1.27× 10 131.6× 10 19(eV)=0.83×106eV=0.83MeV

Conclusion:

The kinetic energy of an electron corresponding to a de Broglie wavelength of 0.01A is 0.83MeV .

(c)

To determine

The energy of an X-ray photon corresponding to a wavelength of 0.01A.

(c)

Expert Solution
Check Mark

Answer to Problem 2.15P

The energy of an X-ray photon corresponding to a wavelength of 0.01A is 1.245MeV .

Explanation of Solution

Given:

Wavelength, λ=0.01A=1×1012m

Formula used:

The energy of a photon is given as:

  E=hcλ

Here,

  h is Planck’s constant

  c is the speed of light

  λ is the wavelength of the photon

Calculation:

The energy of a photon is given as

  E=hcλ

Plugging the values in the above equation

  E=6.64× 10 34×3× 1081× 10 12=1.992×1013J=1.992× 10 131.6× 10 19(eV)=1.245×106eV=1.245MeV

Conclusion:

The energy of an X-ray photon corresponding to a wavelength of 0.01A is 1.245MeV .

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Chapter 2 Solutions

Introduction To Health Physics

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