Atomic Physics

Question 1

Calculate the Kinetic Energy gained by an electron when it is accelerated through a P.D. of

50kV in an X-ray tube• Firstly we know that KE is measured in

Joules.

• JDeV…..

• eV = ½mv2(KE)• eV = (1.6 X 10-19)(50000) = 8 X 10-15 Joules = KE

Calculate the minimum wavelength of an X-ray emitted from the anode.(Planck’s = 6.6 X 10-34Js; c = 3 X 10-8ms-1; Charge on electron = 1.6 X 10-19C)

• What formulae do we know with Planck?

• E = hf

• E = hc/λ

• We know E from last question = 8 X 10-15 J

Calculate the minimum wavelength of an X-ray emitted from the anode.(Planck’s = 6.6 X 10-34Js; c = 3 X 10-8ms-1;

Charge on electron = 1.6 X 10-19C)• E = hc/λ

• 8 X 10-15 J = (6.6 X 10-34)(3 X 10-8ms-1)/λ

• λ = (6.6 X 10-34)(3 X 10-8ms-1)/ 8 X 10-15

• λ = 2.475 X 10-27 m

Question 2

The work function of zinc is 6.9 X 10-19J. What is the minimum frequency of UV

radiation that will cause the photoelectric effect to occur in zinc.

(Planck’s = 6.6 X 10-34Js)

• The main equation that contains work function (and the most important is E = Φ + ½ mv2

• This is also written as:

• hf = hfo + ½ mv2

The work function of zinc is 6.9 X 10-19J. What is the minimum frequency of UV

radiation that will cause the photoelectric effect to occur in zinc. (Planck’s = 6.6 X

10-34Js)

• E = Φ + ½ mv2

• hf = hfo + ½ mv2

• Hence Φ = hfo

• fo = Φ/h = 6.9 X 10-19J/ 6.6 X 10-34Js

• fo = 1.05 X 10-15 Hz

Question 3

Zinc is illuminated with UV light of wavelength 240nm. The work function is 4.3

eV. Calculate the threshold frequency of zinc and Max KE of emitted electrons.

(Planck’s = 6.6 X 10-34Js; c = 3 X 10-8ms-1; Charge on electron = 1.6 X 10-19C)

• What is the main formula we use?• What constants might give us a hint?)• E = Φ + ½ mv2

• hf = hfo + ½ mv2

• Hence Φ = hfo

Zinc is illuminated with UV light of wavelength 240nm. The work function is 4.3

eV. Calculate the threshold frequency of zinc and Max KE of emitted electrons

• Data above has to be converted into forms that we can use.

• 240 nm…………..240 X 10-9 m• The work function 4.3 eV needs to be converted

into Joules……….JDeV………..4.3 eV becomes 4.3 X (1.6 X 10-19) = 6.4 X 10-19 Joules

Zinc is illuminated with UV light of wavelength 240nm. The work function is 4.3

eV. Calculate the threshold frequency of zinc and Max KE of emitted electrons

• E = Φ + ½ mv2

• hf = hfo + ½ mv2

• Hence Φ = hfo

• fo = Φ/h = 6.4 X 10-19J/ 6.6 X 10-34Js

• fo = 0.97 X 10-15 Hz

Zinc is illuminated with UV light of wavelength 240nm. The work function is 4.3

eV. Calculate the threshold frequency of zinc and Max KE of emitted electrons

• E = Φ + ½ mv2

• hc/λ = hfo + ½ mv2

• (6.6 X 10-34)(3 X 10-8) = 6.4 X 10-19 + ½ mv2

240 X 10-9

• (6.6 X 10-34)(3 X 10-8) = ½ mv2 (KE) (240 X 10-9)(6.4 X 10-19)

Question 4

An electron accelerates from the cathode through a P.D. of 4kV in a CRT.

How much energy does the electron gain? What is the speed of the electron at the

anode?

• As usual we see that they are looking for the energy gained

• eV = ½mv2

• eV = (1.6 X 10-19)(4000) = 6.4 X 10-16 Joules = KE

An electron accelerates from the cathode through a P.D. of 4kV in a CRT.

How much energy does the electron gain? What is the speed of the electron at the

anode?

• eV = ½mv2

• 6.4 X 10-16 Joules = ½mv2

• 6.4 X 10-16 Joules = ½(9.1 X 10-31)v2

• (6.4 X 10-16)(2) = v2

(9.1 X 10-31)

• V = 1.18 X 108 ms-1

After leaving the anode, the electron travels at a constant speed and enters a magnetic field at right angles where it is reflected. The flux density is 5 X 10-2

T. Calculate the force acting on the electron?

• F = qvB

• F = (1.6 X 10-19)(1.18 X 108)(5 X 10-2)

• F = 9.44 X 10-13 N

After leaving the anode, the electron travels at a constant speed and enters a magnetic

field at right angles where it is reflected. The flux density is 5 X 10-2 T. Calculate the

radius of the circular path followed by the electron in the field?

• F = qvB F = mv2/r• qvB = mv2/r• qvBr = mv2

• qBr = mv• r = mv/qB

After leaving the anode, the electron travels at a constant speed and enters a magnetic

field at right angles where it is reflected. The flux density is 5 X 10-2 T. Calculate the

radius of the circular path followed by the electron in the field?

• r = mv/qB

• r = (9.1 X 10-31)(1.18 X 108)

(1.6 X 10-19)(5 X 10-2)

r = 1.34 X 10-2 m

Question 5

Radium-226 undergoes α-decay and has a decay constant of 1.35 X 10-11s-1.

Calculate the number of α-particles emitted per second by a 2μg sample of

this isotope.(1 mol of radium-226 is 226 g; Avagadros constant

= 6.02 X 1021 mol-1)

• 1 mol = 226 g = 6.02 X 1021 atoms

• 1g =6.02 X 1021/226 = 2.66 X 1019 atoms

• 1μg = 2.66 X 1019 / 106= 2.66 X 1013 atoms• 2μg = 5.32 X 1013 atoms

Radium-226 undergoes α-decay and has a decay constant of 1.35 X 10-11s-1.

Calculate the number of α-particles emitted per second by a 2μg sample of

this isotope.(1 mol of radium-226 is 226 g; Avagadros constant

= 6.02 X 1021 mol-1)

• Number of α-particles = λN

• λN = (5.32 X 1013)(1.35 X 10-11)

• No. of α-particles = λN = 7.182 X 102 particles

Question 6

A detector records 1200 counts per minute when the activity of a

radioactive sample is first measured. Six minutes later the activity has fallen to 150 counts per minute. Calculate the

half-life of the sample

• 1200 600 300 150

1 half-life 2 half-lives 3 half-lives

• 3 half lives in 6 min…Each half-life in 2 min

Question 7

An ancient wooden cup from an archaeological site has an activity of 2.1 Bq. The corresponding for newly cut wood is 8.4 Bq. If the half-life of

Carbon-14 is 5730 years, estimate the age of the cup.

• 8.4Bq 4.2Bq 2.1Bq 1 half-life 2 half-lives• 2 half lives each half-life being 5730 years

each…….11460 years

Question 8

The power generator in a nuclear reactor is 150MW. Calculate the number

of fissions occurring per second in the reactor, give that 180MeV of energy is

released per fission• Firstly, get all units into SI units.

• 150 MW is 150 X 106 Watts

• 180MeV is 180 X 106 eV but this isn’t an SI• 180 x 106 eV…….…..…JDeV……………..

(180 x 106)(1.6 X 10-19) = 2.88 X 10-11 Joules

The power generator in a nuclear reactor is 150MW. Calculate the number of fissions

occurring per second in the reactor, give that 180MeV of energy is released per fission

• Watts = 150 X 106 Watts• Watts = Energy per Second• Energy released per fission = 2.88 X 10-11 J

• Fissions per Sec = Energy/ Energy released per fission

• Fissions per Sec = (150 X 106)/(2.88 X 10-11)

= 4.32 X 10-3 fissions per sec

Question 9

If a sample of radium contains 2.6 X 1021 radium-226 and is emitting 3.5 X

1010 particles per second, calculate the decay constant

• Rate of decay = λN

• Rate of decay/N = λ• 3.5 X 1010/ 2.6 X 1021 = λ• λ = 1.35 X 10-11 s-1

If a sample of radium contains 2.6 X 1021 radium-226 and is emitting 3.5 X 1010

particles per second, calculate the half-life if the rate of decay is 1.35 X 10-11 s-1

• T½ = 0.693/λ

• T½ = 0.693/1.35 X 10-11

• T½ = 5.15 X 1010 s

Question 10

A large number of solar panels are joined together in series and cover an area of 20m2. The efficiency of the solar cells is 20%. If the solar cell constant is 1400

Wm-2, what is the max power generated by the solar cells

• Because the efficiency is only 20% on 20m2, it’s like having only 4m2 solar cells.

• Solar constant is 1400 Wm-2. From the unit I see that this is power/ area

A large number of solar panels are joined together in series and cover an area of 20m2. The efficiency of the solar cells is 20%. If the solar cell constant is 1400

Wm-2, what is the max power generated by the solar cells

• Solar constant is 1400 Wm-2. From the unit I see that this is power/ area.

• 1400 = Power/ area• 1400 X area = Power • 1400 X 4 = Power = 5600W

Question 11

Cobalt-60 is a radioactive isotope with a half-life of 5.26 years and

emits β-particles. Write an equation to represent the decay of cobalt-60.• Cobalt-60 ---------- β + substance-61

Cobalt-60 is a radioactive isotope with a half-life of 5.26 years and emits β-particles. Calculate the

decay constant of cobalt-60.

• T½ = 0.693/λ

• λ = 0.693/ T½

• λ = 0.693/5.26

• λ = 0.13175 y-1

Cobalt-60 is a radioactive isotope with a half-life of 5.26 years and emits β-

particles. Calculate the rate of decay of cobalt-60 when it has 2.5 X 1021 atoms.

Rate of decay = λNRate of decay = λ X NRate of decay = 0.13175/ 2.5 X 1021

Rate of decay = 0.33 X 1021

Rate of decay = 3.3 X 1020 particles

Question 12

A neutral pion is unstable with a decay constant of 2.5 X 1012 s-1. What is the

half-life of a neutral pion?

• T½ = 0.693/λ

• T½ = 0.693/2.5 X 1012

• T½ = 2.772 X 10-13 s