The number of neutrons in^6_3 Li is a. 9 b. 8 c. 6 d. 3 e. 2 Uranium-235 has a h
ID: 517524 • Letter: T
Question
The number of neutrons in^6_3 Li is a. 9 b. 8 c. 6 d. 3 e. 2 Uranium-235 has a half-life of 7.04 times 10^8 years. How many years will it take for 99.9% of a U-235 sample to decay? a. 7.0 times 10^5 yr b. 1.0 times 10^6 yr c. 4.7 times 10^9 yr d. 4.9 times 10^9 yr e. 7.0 times 10^9 yr When the radioactive nuclide^199At undergoes alpha emission, what is the product nuclide? a.^197Bi b.^203Fr c.^197Rn d.^195Bi e.^199Fr If a tree dies and the trunk remains undisturbed for 14996 years, what percentage of original^14C is still present? (half-life of^14C = 5730 years) a. 32.0% b. 24.0%. c. 83.0% d. 16.0% An atom with 45 protons has a mass number of 100. It must contain how many neutrons? a. 145 b 45 c. 100 d 55 e. none of theseExplanation / Answer
51. d. 3 [ here mass number = no pf proton + no of neutron = 6 and number of proton (atomic mass) = 3 so number of neutron = 3]
52. e
the decay will first order kinetics
Given, half life, t1/2 = 7.04 x 108 years so k = 0.693/t1/2 = 0.693 / (7.04 x 108) years-1
t = (2.303/k)log (N0/N) where N0 = initial amount and N = final amount ; k = rate constant
= {(2.303 x 7.04 x 108)/0.693} log (100/0.1) N0 = 100 ; N = 100- 99.9 = 0.1
= (23.4 x 108) x 3 = 7.02 x 109 years
53. d. 195Bi
199 At ------> 24He + 195Bi on emission of alpha radiation the mass number reduces by 4 units.
54. d. 16%
the decay will first order kinetics
Given, half life, t1/2 = 5730 years so k = 0.693/t1/2 = 0.693 / 5730 years-1
t = 14996 years
t = (2.303/k)log (N0/N) where N0 = initial amount and N = final amount ; k = rate constant
14996 = {(2.303 x 5730)/0.693} log (N0/N)
log (N0/N) = (14996 x 0.693)/ ( (2.303 x 5730) = 0.788
(N0/N) = antilog (0.788) = 6.14
N/N0 = 0.16
N = 0.16 N0
If N0 = 100 then N = 16%
55. d. 55 no of neutron = mass number - no or proton = 100- 45 = 55