The age of most ancient geological formations is estimated by
Emission of an alpha particle leads to a
[1989]
Emission of α particle ) leads to
decrease of 2 units of charge. e.g
If an isotope of hydrogen has two neutrons in itsatom, its atomic number and atomic mass numberwill respectively be [1992]
As number of neutron = Mass number
– atomic number
Give number of neutron = 2
∴ Mass number will be 3 and atomic number
will be one.
In a radioactive decay, an emitted electron comesfrom [1994]
When a radioactive elements emits α or β
particle the new element formed may have
unstable nucleus. It may further disintegrate
by emitting α or β particle forming a new
element. This process of integration may
continue till end product formed is a stable
compound.
India has the world’s largest deposits of thoriumin the form of [1994]
The ore of thorium is monazite.
Halflife for radioactive 14C is 5760 years. In howmany years, 200 mg of 14C will be reduced to 25mg? [1995]
Halflife of ^{14}C = 5760 yrs; Initial weight of
^{14}C = 200 mg and final weight of ^{14}C = 25 mg.
Quantity left after 5760 years = 200/2 = 100 mg
Similarly quantity left after another 5760
years (i.e 11520 years) = 100/2 = 50 mg
Quantity left after another 5760 years
(i.e. 17280 years) = 50/2 = 25 mg
Thus time taken by 200 mg of 14C to reduce
to 25 mg = (5760 + 5760 + 5760 ) years = 17280
years.
Alternative solution
As we know that
where N_{0} original amount of radioactive
sustacnce
N = Amount of substance remain after n half
lives
where T = total time
T = 3 × 5760 years = 17280 years
One microgram of radioactive sodium ^{24}_{11}Na witha halflife of 15 hours was injected into a livingsystem for a bioassay. How long will it take forthe radioactive subtance to fall up to 25% of theinitial value? [1996]
Carbon  14 dating method is based on the factthat: [1997]
By carbon dating method
Hence it is based upon the ratio of C^{14} and
C^{12}.
fission products + neutrons + 3.20 × 10^{–11} J The energy released when 1 g of finally undergoes fission is
1 atom of on fission gives energy
= 3.2 × 10^{–11} J
6.023 × 1023 atom (1 mole) on fission gives
energy = 3.2 × 10^{–11} × 6.023 × 10^{23} J
235 gm of on fission gives energy
Number of neutrons in a parent nucleus X, which gives nucleus, after two successive β emissions, would be
No. of neutrons = Mass number – No. of
proton =14 – 5 = 9
When a radioactive element emits successivelyone αparticle and two βparticles, the massnumber of the daughter element [1999]
Mass number is effected by emmision of α
particle while β particle has negligible mass
does not effect mass number. e.g
A human body required 0.01M activity ofradioactive substance after 24 hours. Half life ofradioactive substance is 6 hours. Then injectionof maximum activity of radioactive substance thatcan be injected will be [2001]
Remaining activity = 0.01M
after 24 hrs
Initial activity = 0.01 × 16 = 0.16M
If species emits firstly a positron, then two α and two β and in last one α and finally converted to species , so correct relation is
nucleus absorbs a neutron and disintegrates into and x. So what will be the product x? [2002]
i.e 3 neutrons
The radioactive isotope, tritium, has a halflifeof 12.3 years. If the initial amount of tritium is32 mg, how many milligrams of it would remainafter 49.2 years? [2003]
Given t_{1/2} = 12.3 years
Initial amount (N0) = 32 mg
Total time = 49.2 years
Hence 32 mg becomes 2 mg in 49.2 years
The radioactive isotope which is used in the treatment of cancer can be made by (n, p) reaction. For this reaction the target nucleus is
Balancing the mass and atomic numbers on
both sides
Thus X should be
A nuclide of an alkaline earth metal undergoesradioactive decay by emission of the αparticles in succession. The group of theperiodic table to which the resulting daughterelement would belong is [2005]
When IIA group element (Ra) emits one
αparticle its group no. decreases by two
unit. i.e., go into zero group (Gr. 16) But as
it is radioactive thus due to successive
emission last product is Pb i.e., (Gr.14).
The half life of a substance in a certain enzymecatalysedreaction is 138s. The time required forthe concentration of the substance to fall from1.28 mg L^{–1} to 0.04 mg L^{–1}, is : [2011]
For a first order reaction
Total time T = no. of half lives (n) × half life
(t_{1/2})
where n = no. of half lives
Give N0 (original amount) = 1.28 mg/ ℓ
N (amount of substance left after time T)
= 0.04 m/g l
n = 5
T = 5 × 138
= 690








