Test: Nuclear Chemistry - From Past 28 Years Questions


18 Questions MCQ Test Chemistry 28 Years Past year papers for NEET/AIPMT Class 12 | Test: Nuclear Chemistry - From Past 28 Years Questions


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QUESTION: 1

The age of most ancient geological formations is estimated by

Solution:
QUESTION: 2

Emission of an alpha particle leads to a           

[1989]

Solution:

Emission of α -particle ) leads to
decrease of 2 units of charge. e.g

QUESTION: 3

If an isotope of hydrogen has two neutrons in itsatom, its atomic number and atomic mass numberwill respectively be    [1992]

Solution:

As number of neutron = Mass number
– atomic number
Give number of neutron = 2
∴ Mass number will be 3 and atomic number
will be one.

QUESTION: 4

In a radioactive decay, an emitted electron comesfrom [1994]

Solution:

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.

QUESTION: 5

India has the world’s largest deposits of thoriumin the form of          [1994]

Solution:

The ore of thorium is monazite.

QUESTION: 6

Half-life for radioactive 14C is 5760 years. In howmany years, 200 mg of 14C will be reduced to 25mg?           [1995]

Solution:

Half-life of 14C = 5760 yrs; Initial weight of
14C = 200 mg and final weight of 14C = 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 N0 original amount of radioactive
sustacnce
N = Amount of substance remain after n half
lives

where T = total time
T = 3 × 5760 years = 17280 years

QUESTION: 7

One microgram of radioactive sodium  2411Na witha half-life of 15 hours was injected into a livingsystem for a bio-assay. How long will it take forthe radioactive subtance to fall up to 25% of theinitial value? [1996]

Solution:

QUESTION: 8

Carbon - 14 dating method is based on the factthat: [1997]

Solution:

By carbon dating method

Hence it is based upon the ratio of C14 and
C12.

QUESTION: 9

fission products + neutrons + 3.20 × 10–11 J The energy released when 1 g of     finally undergoes fission is

Solution:

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 × 1023 J
235 gm of  on fission gives energy 

QUESTION: 10

Number of neutrons in a parent nucleus X, which  gives    nucleus, after two successive β emissions, would be

Solution:

No. of neutrons = Mass number – No. of
proton =14 – 5 = 9

QUESTION: 11

When a radioactive element emits successivelyone α-particle and two β-particles, the massnumber of the daughter element [1999]

Solution:

Mass number is effected by emmision of α
particle while β particle has negligible mass
does not effect mass number. e.g

QUESTION: 12

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]

Solution:

Remaining activity = 0.01M
after 24 hrs

 

Initial activity = 0.01 × 16 = 0.16M

QUESTION: 13

 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

Solution:

QUESTION: 14

 nucleus absorbs a neutron and disintegrates into   and x. So what will be the product x? [2002]
 

Solution:

i.e 3 neutrons

QUESTION: 15

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]

Solution:

Given t1/2 = 12.3 years
Initial amount (N0) = 32 mg
Total time = 49.2 years

Hence 32 mg becomes 2 mg in 49.2 years

QUESTION: 16

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

Solution:

Balancing the mass and atomic numbers on
both sides

Thus X should be 

QUESTION: 17

A nuclide of an alkaline earth metal undergoesradioactive decay by emission of the α-particles in succession. The group of theperiodic table to which the resulting daughterelement would belong is [2005]

Solution:

When IIA group element (Ra) emits one
α-particle 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).

QUESTION: 18

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]

Solution:

For a first order reaction
Total time T = no. of half lives (n) × half life
(t1/2)

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