17.A scintillation counter A) measures the signal coming from an ionized gas. B)

Question

17.A scintillation counter A) measures the signal coming from an ionized gas. B) measures light emissions from excited atoms. C) depends on an avalanche ofelectrons generated as a particle moves through a tube of argon gas. D) detects high energy radiation better than low energy radiation. E) detects an electric current in a gas. 18 The isotope Mg has a half-ife of21 hours. Ifa sample initially contains exactly 10000 atoms of aMg approximately how many of these atoms will remain after one week? A) 1250 B) 78 C) 39 D) 0 E) none of the above

Explanation / Answer

17) A scintillator counter measures ionization radiation by exciting a charged particle and emitting photons. These photons strike a photomultiplier tube’s photocathode, leading to a photoelectric effect and hence generation of current. Thus, (B) is the most appropriate option.

Ans: (B)

18) 1 week = 7 days; 1 day = 24 h; therefore, 1 week = (7 days)*(24 h/1 day) = 168 h.

Use the radio-active decay law:

ln Nt/N0 = -k*t where k = 0.693/t1/2 where t1/2 = 21 h is the half life of the radio element. Plug in values and write

ln Nt/N0 = -(0.693/21 h)*(168 h) = -5.544

===> Nt/N0 = exp(-5.544) = 0.00391

Given N0 = 10,000 atoms, we have,

Nt/(10,000 atoms) = 0.00391

===> Nt = 0.00391*(10,000 atoms) = 39.1 39

Ans: (C)

19) We will follow the radioactive decay law. 0.150 mole silver-107 is formed from 1.25 mole Pd-107; therefore, amount of Pd-107 left = (1.25 – 0.150) mole = 1.10 mole; thus, N0 = 1.25 mole, Nt = 1.10 mole.

Use the radio-active decay law as noted above.

ln Nt/N0 = -(0.693/6.5*105 yr)*t

===> ln (1.10 mole/1.25 mole) = -(0.693/6.5*105 yr)*t

===> -0.1278 = -(0.693/6.5*105 yr)*t

===> t = (0.1278*6.5*105 yr)/(0.693) = 1.198*105 yr 1.2*105 yr

Ans: (E)

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