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What are X-rays? Electrons are produced and used in an X-ray tube - Leaving Cert Physics - Question 10 - 2017

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What are X-rays? Electrons are produced and used in an X-ray tube. How are the electrons produced? Where in the tube does this take place? (12) A certain X-ray tub... show full transcript

Worked Solution & Example Answer:What are X-rays? Electrons are produced and used in an X-ray tube - Leaving Cert Physics - Question 10 - 2017

Step 1

How are the electrons produced? Where in the tube does this take place?

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Answer

Electrons in an X-ray tube are produced through a process called thermionic emission. This occurs in the cathode or filament of the tube, where the filament is heated to a high temperature, allowing electrons to escape from the surface.

Step 2

Calculate (i) the energy of an X-ray photon of wavelength 0.02 nm

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Answer

The energy of an X-ray photon can be calculated using the formula:

E=hfE = hf

where:

  • hh is Planck's constant, approximately 6.626imes1034extJs6.626 imes 10^{-34} ext{ Js},
  • ff is the frequency calculated using the speed of light cc and the wavelength extλ ext{λ}:

f = rac{c}{ ext{λ}}

Using the wavelength extλ=0.02extnm=0.02imes109extm ext{λ} = 0.02 ext{ nm} = 0.02 imes 10^{-9} ext{ m}:

f = rac{3.00 imes 10^8 ext{ m/s}}{0.02 imes 10^{-9} ext{ m}} = 1.5 imes 10^{16} ext{ Hz}

Now substituting for EE:

E=(6.626imes1034extJs)(1.5imes1016extHz)E = (6.626 imes 10^{-34} ext{ Js})(1.5 imes 10^{16} ext{ Hz})

Calculating gives: Eextext9.9imes1015extJ.E ext{ } ext{≈ } 9.9 imes 10^{-15} ext{ J}.

Step 3

Calculate (ii) the maximum velocity of an electron in the tube

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Answer

The maximum velocity can be calculated using the kinetic energy formula:

E=12mv2E = \frac{1}{2} mv^2

Rearranging gives:

v=2Emv = \sqrt{\frac{2E}{m}}

Where:

  • EE is the energy calculated previously (9.9imes1015extJ9.9 imes 10^{-15} ext{ J})
  • mm is the mass of an electron, approximately 9.11imes1031extkg9.11 imes 10^{-31} ext{ kg}.

Thus,

v=2(9.9imes1015)9.11imes1031extm/sv = \sqrt{\frac{2(9.9 imes 10^{-15})}{9.11 imes 10^{-31}}} ext{ m/s}

Calculating yields: vextext1.48imes108extm/s.v ext{ } ext{≈ } 1.48 imes 10^8 ext{ m/s}.

Step 4

Calculate (iii) the voltage applied to the electrons

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Answer

The voltage applied to the electrons can be calculated from the relationship between energy and charge:

E=qVE = qV

Rearranging gives:

V=EqV = \frac{E}{q}

For an electron, the charge qextext1.6imes1019extCq ext{ } ext{≈ } 1.6 imes 10^{-19} ext{ C}.

Substituting:

V=9.9imes10151.6imes1019V = \frac{9.9 imes 10^{-15}}{1.6 imes 10^{-19}}

Calculating yields: Vextext62000extV.V ext{ } ext{≈ } 62000 ext{ V}.

Step 5

Draw a labelled diagram of a photocell

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Answer

The labelled diagram of a photocell includes the following components:

  • Anode
  • Semi-cylindrical cathode
  • Case
  • Vacuum Any additional components can be labelled as necessary.

Step 6

Describe how a photocell conducts current.

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Answer

A photocell conducts current by allowing light of suitable frequency to strike the cathode, which emits electrons. The emitted electrons move toward the anode, generating a flow of current as they gain energy from the light.

Step 7

How many electrons are generated in the photocell during each minute?

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Answer

Given a current of 2extµA=2imes106extA2 ext{ µA} = 2 imes 10^{-6} ext{ A}, using the equation:

Q=ItQ = It

where:

  • QQ is charge,
  • II is current,
  • tt is time in seconds (1 minute = 60 seconds):

Q=(2imes106)(60)=1.2imes104extCQ = (2 imes 10^{-6})(60) = 1.2 imes 10^{-4} ext{ C}

Now, to find the number of electrons:

n=Qen = \frac{Q}{e}

Where:

  • e=1.6imes1019extCe = 1.6 imes 10^{-19} ext{ C} is the charge of one electron.

Thus,

n=1.2imes1041.6imes1019=7.5imes1014extelectrons.n = \frac{1.2 imes 10^{-4}}{1.6 imes 10^{-19}} = 7.5 imes 10^{14} ext{ electrons}.

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