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JEE Main > Nuclear Chemistry

Explore popular questions from Nuclear Chemistry for JEE Main. This collection covers Nuclear Chemistry previous year JEE Main questions hand picked by experienced teachers.

Q 1.

Correct4

Incorrect-1

Sulphur-35 (34.96903 amu) emits a β-particle but no γ-ray. The product is chlorine-35 (34.96885 amu). The maximum energy emitted by the β-particle is

A

16.758 MeV

B

1.6758 MeV

0.16758 MeV

D

0.016758 MeV

Explanation

The mass converted into energy

()
Energy produced

Q 2.

Correct4

Incorrect-1

If the atomic masses of Lithium, Helium and Proton are 7.01823 amu, 4.00387 amu and 1.00815 amu respectively, calculate the energy that will be evolved in the reaction. .
(Given that 1 amu = 931 MeV)

17.3 MeV

B

17.8 MeV

C

17.2 MeV

D

17.0 MeV

Explanation

Total mass of the reacting species
and
The mass of the resulting species
amu
Mass of reacting species converted into energy, i.e., amu
Energy evolved in the reaction
MeV.

Q 3.

Correct4

Incorrect-1

Calculate the mass defect and binding energy per nucleon for . [The mass of , mass of hydrogen atom = 1.008142 amu and mass of neutron = 1.008982 amu].

8.77 MeV

B

8.25 MeV

C

9.01 MeV

D

8.00 MeV

Explanation

Number of protons in
Number of neutrons = 59 - 27 = 32
The binding energy per nucleon

Q 4.

Correct4

Incorrect-1

Calculate the number of neutrons in the remaining atom after emission of an alpha particle from atom

A

146

B

145

144

D

143

Explanation

On account of emission of an alpha particle, the atomic mass is decreased by 4 units and atomic number by 2 units
So, Atomic mass of daughter element = 234
Atomic number of daughter element = 90
Number of neutrons = Atomic mass - Atomic number
= 234 - 90 = 144

Q 5.

Correct4

Incorrect-1

Radioactive disintegration of takes place in the following manner into
, Determine mass number and atomic number of RaC.

A

214 and 84

B

214 and 86

214 and 83

D

214 and 85

Explanation

Parent element is
Atomic mass = 226
Atomic number = 88
RaC is formed after the emission of 3 alpha particles. Mass of 3 alpha particles
So Atomic mass of RaC
With emission of one α-particle, atomic number is decreased by 2 and with emission of β-particle, atomic number is increased by 1.
So Atomic number of RaC

Q 6.

Correct4

Incorrect-1

How many 'α and β' particles will be emitted when changes into

A

2 and 6

4 and 2

C

2 and 4

D

6 and 2

Explanation

The change is;
Decrease in mass
Mass of 1 α-particle
Therefore, number of α-particles emitted
Number of β-particles emitted
Hence number of α-particles = 4 and number of β-particles = 2

Q 7.

Correct4

Incorrect-1

An atom has atomic mass 232 and atomic number 90. During the course of disintegration, it emits 2 β-particles and few α-particles. The resultant atom has atomic mass 212 and atomic number 82. How many α-particles are emitted during this process

5

B

6

C

7

D

8

Explanation

The decrease in atomic mass = (232 - 212) = 20
Decrease in mass occurs due to emission of α-particles. Let x be the number of alpha particles emitted.
Mass of 'x' α-particles = 4x
So or
Alternative method: This can also be determined by the application of following equation:
No. of β-particles emitted = 2 × No. of α-particles emitted
2 = 2 × x - (90 - 82) or x = 5

Q 8.

Correct4

Incorrect-1

An element X with atomic number 90 and mass number 232 loses one α- and two β-particles successively to give a stable species Z. What would be the atomic number and atomic weight of Z

B

C

D

Explanation

At. no. and At. wt. of the element (Y) produced by the loss of
One α-particle from
At. no. and At.wt. of Z produced by the loss of 2 β-particles
from

Q 9.

Correct4

Incorrect-1

Find out the total number of α- and β-particles emitted in the disintegration of to

6 and 4

B

8 and 4

C

9 and 6

D

2 and 4

Explanation

Change in at. wt. = 232 - 208 = 24 amu (at. wt. unit)
Now since in one α-particle emission, at. wt. is decreased by
4 amu, the number of α-emissions for 24 amu = 24/4 = 6
Atomic number after 6α-emissions = 90 - 12 = 78

Increase in atomic number from 78 to the given 82 = 82 - 78
= 4
No. of β-particle emissions = 4

Q 10.

Correct4

Incorrect-1

disintegrates to give as the final product. How many alpha and beta particles are emitted in this process

6

B

8

C

9

D

2

Explanation


Decrease in mass = (234 - 206) = 28
Mass of α-particle = 4
So Number of α-particles emitted
Number of beta particles emitted = 2 × No. of α-particles - (At. no. of parent - At. no. of end product)
= 2 × 7 - (90 - 82) = 6

Q 11.

Correct4

Incorrect-1

The isotopes and occur in nature in the ratio of 140 : 1. Assuming that at the time of earth formation, they were present in equal ratio, make an estimation of the age of earth. The half life period of and are and years respectively

years

B

years

C

years

D

years

Explanation

Let the age of the earth be t years
For ......(i)
For ......(ii) Subtracting eq. (ii) from eq. (i)



years

Q 12.

Correct4

Incorrect-1

Calculate the mass of (half life period = 5720 years) atoms which give disintegrations per second

A


C

D

Explanation

Let the mass of atoms be m g
Number of atoms in m g of
We know that
i.e. Rate of disintegration

Q 13.

Correct4

Incorrect-1

A certain radio-isotope (Half life = 10 days) decays to . If 1 g of atoms of is kept in sealed vessel, how much helium will accumulate in 20 days

16800 Ml

B

17800 mL

C

18800 mL

D

15800 mL

Explanation


In two half lives, of the isotope has disintegrated, i.e., g atom of helium has been formed from g atom of
Volume of 1 g atom of helium = 22400 Ml
Thus, Volume of g atom of helium

Q 14.

Correct4

Incorrect-1

In nature a decay chain series starts with and finally terminates at A thorium ore sample was found to contain of helium at STP and of Find the age of the ore sample assuming the source of helium to be only decay of . Also assume complete retention of helium within the ore. (Half life of years)

A

years

years

C

years

D

years

Explanation

No. of moles of helium

No. of moles which have disintegrated
Mass of which have disintegrated

Mass of left,

Applying years

Q 15.

Correct4

Incorrect-1

Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron

80

B

90

C

70

D

20

Explanation

Let the % of isotope with at. wt. 10.01 = x
% of isotope with at. wt. 11.01 = (100 - x)
Now since,


Hence, % of isotope with at. wt. 10.01 = 20
% of isotope with at. wt. 11.01 = 100 - 20 = 80

Q 16.

Correct4

Incorrect-1

An element of avg. atomic mass z, has a isotopic masses (z + 2) and (z - 1). Therefore percentage abundance of the heavier isotope is -

%

B

%

C

%

D

× 100%

Explanation

Let % abundance of heavier isotope is x %
= z
or xz + 2x - xz + x + 100z - 100 = 100z
or 3x = 100 or x = %

Q 17.

Correct4

Incorrect-1

m gram of a radioactive species (at. Mass M) has decay constant λ. The initial specific activity at zero time is given by

λ .

B

λ .

C

D

m Meλ

Explanation

Specific activity = rate per g or ds-1g-1
= λ . = λ .

Q 18.

Correct4

Incorrect-1

In the chain reaction involving fission of as

Neutrons and energy produced at the nth step is-

A

3n ,nE

B

3n , nE

3n , 3n -1 E

D

3n -1 , nE

Explanation

In nth steps, neutrons ejected are 3n and En -1 because one step ejects 3 neutrons and energy E.

Q 19.

Correct4

Incorrect-1

At the point of intersection of the two curves shown, the concentration of B is given by ........... for, A ―→nB :

A

B

D

Explanation

[A]left = [B]formed = n × [A]decayed
A0 e-λt = n × A0 (1 - e-λt)
e-λt =
[B]formed = n × A0 × (1 - e-λt)
= n × A0 ×
=

Q 20.

Correct4

Incorrect-1

A G.M. counter is used to study the radioactive process of first-order. In the absence of radioactive substance A, it counts 3 disintegrations per second (dps). When A is placed in the G.M. counter, it records 23 dps at the start and 13 dps after 10 minutes. It records x dps after next 10 minutes and A has half-life period y minutes. x and y are :

8 dps, 10 min

B

5 dps, 10 min

C

5 dps, 20 min

D

5 dps, 5 min

Explanation

There is an error of 3 dps
C0 = 20 dps
Ct = 10 dps
Thus, half - life = 10 minutes
In next 10 minutes Ct = 5 dps
recorded value with error = 8 dps