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
16.758 MeV
1.6758 MeV
0.16758 MeV
0.016758 MeV
The mass converted into energy
()
Energy produced
Correct4
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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
17.8 MeV
17.2 MeV
17.0 MeV
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 reactionMeV.
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
8.25 MeV
9.01 MeV
8.00 MeV
Number of protons in
Number of neutrons = 59 - 27 = 32The binding energy
per nucleon
Correct4
Incorrect-1
Calculate the number of neutrons in the remaining atom after emission of an alpha particle from atom
146
145
144
143
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
Correct4
Incorrect-1
Radioactive disintegration of takes place in the following manner into
, Determine mass number and atomic number of RaC.
214 and 84
214 and 86
214 and 83
214 and 85
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
Correct4
Incorrect-1
How many 'α and β' particles will be emitted when changes into
2 and 6
4 and 2
2 and 4
6 and 2
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
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
6
7
8
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
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
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
Correct4
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Find out the total number of α- and β-particles emitted in the disintegration of to
6 and 4
8 and 4
9 and 6
2 and 4
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
Correct4
Incorrect-1
disintegrates to give
as the final product. How many alpha and beta particles are emitted in this process
6
8
9
2
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
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
years
years
years
Let the age of the earth be t years
For
......(i)
For
......(ii) Subtracting eq. (ii) from eq. (i)
years
Correct4
Incorrect-1
Calculate the mass of (half life period = 5720 years) atoms which give
disintegrations per second
Let the mass of atoms be m g
Number of atoms in m g ofWe know that
i.e. Rate of disintegration
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
17800 mL
18800 mL
15800 mL
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
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)
years
years
years
years
No. of moles of helium
No. of moles which have disintegrated
Mass of which have disintegrated
Mass of left,
Applying years
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
90
70
20
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
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 -
%
%
%
× 100%
Let % abundance of heavier isotope is x %
= z
or xz + 2x - xz + x + 100z - 100 = 100z
or 3x = 100 or x = %
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 .
λ .
m Meλ
Specific activity = rate per g or ds-1g-1
= λ . = λ .
Correct4
Incorrect-1
In the chain reaction involving fission of as
Neutrons and energy produced at the nth step is-
3n ,nE
3n , nE
3n , 3n -1 E
3n -1 , nE
In nth steps, neutrons ejected are 3n and En -1 because one step ejects 3 neutrons and energy E.
Correct4
Incorrect-1
At the point of intersection of the two curves shown, the concentration of B is given by ........... for, A ―→nB :
[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 ×
=
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
5 dps, 10 min
5 dps, 20 min
5 dps, 5 min
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