The activity of a radioactive sample is measuredas 9750 counts per minute at t = 0 and as 975counts per minute at t = 5 minutes. The decayconstant is approximately [1997]
9750 = KN_{0} ............. (1)
975 = KN ............. (2)
Dividing (1) by (2)
= 0.4606 = 0.461 per minute
The stable nucleus that has a radius half that of Fe^{56} is [1997]
The nuclear radius R = R_{o}A^{1/3}
In a fission reaction
the binding energy per nucleon of X and Y is 8.5MeV whereas of ^{236}U is 7.6 MeV. The totalenergy liberated will be about [1997]
Binding energy
= 117 × 8.5 + 117 × 8.5 – 236 × 7.6
= 234 × 8.5 – 236 × 7.6
= 1989 – 1793.6 = 200 MeV
Thus, in per fission of Uranium nearly 200
MeV energy is liberated
A free neutron decays into a proton, an electronand [1997]
X must have zero charge and almost zero mass as electron is emitted. Hence X must be antineutrino.
The most penetrating radiation of the following is [1997]
The penetrating power of radiation is directly proportional to the energy of its photon.
Energy of a photon
∴ Penetrating power
λ is minimum for γrays, so penetrating power is maximum of γrays.
Which of the following is used as a moderator innuclear reactors? [1997]
Moderator used in nuclear r eactor are graphite and heavy water.
Halflives of two radioactive substances A and B are respectively 20 minutes and 40 minutes.Initially, the samples of A and B have equalnumber of nuclei. After 80 minutes the ratio ofremaining numbers of A and B nuclei is [1998]
80 = 20 × n_{A} ⇒ n_{A} = 4
80 = 40 × n_{B} ⇒ n_{B} = 2
Atomic weight of Boron is 10.81 and it has two isotopes . Then the ratio in nature would be [1998]
Suppose that, The number of ^{10}B type atoms = x and the number of ^{11}B type atoms = y
Weight of ^{10}B type atoms = 10x
Weight of ^{11}B type atoms = 11y
Total number of atoms = x + y
∴ Atomic weight
⇒10x + 11y = 10.81x + 10.81y
⇒ 0.81x = 0.19y
A nucleus _{n}X^{m} emits one α and two β particles. The resulting nucleus is
Complete the equation for the following fission process : [1998]
Alphaparticles are [1999]
We know that alpha particles are the nucleus of ionized helium atoms which contain two protons and two neutrons. These are emitted by the nuclei of certain radioactive substances. Streams of alpha particles, called αrays, produce intense ionisation in gases through which they pass and are easily absorbed by matter.
After 1α and 2 βemissions [1999]
Emission of 1α particle led to decrease in atomic number by 2 while mass number by 4.
On the other hand, emission of 2β particles increases atomic number by 2. Hence, overall emission of, 1α and 2β particles led to decrease in mass number by 4.
The decay constant (λ) and the halflife (T) of a radioactive isotope are related as [2000]
when
Atomic hydrogen has life period of [2000]
Atomic hydrogen is unstable and it has life period of a fraction of a second.
It is possible to understand nuclear fission onthe basis of the [2000]
According to liquid drop model of nucleus, an excited nucleus breaks into lighter nuclei just like an excited drop breaks into tiny drops.
In the following nuclear reaction
what does X stand for? [2000]
X is a neutrino, when βparticle is emitted.
Mn and Mp represent mass of neutron and protonrespectively. If an element having atomic massM has Nneutron and Zproton, then the correctrelation will be [2001]
Given : Mass of neutron = M_{n} Mass of proton = M_{p}; Atomic mass of the element = M ;
Number of neutrons in the element = N and number of protons in the element = Z.
We know that the atomic mass (M) of any stable nucleus is always less than the sum of the masses of the constituent particles.
Therefore, M < [NM_{n} + ZM_{p}].
A sample has 4 × 10^{16} radioactive nuclei of halflife 10 days. The number of atoms decaying in 30 days is [2002]
Atoms decayed
= 3.5 × 10^{16}
A deuteron strikes _{8}O^{16} nucleus with subsequent emission of an alpha particle. Identify the nucleus so produced [2002]
For a nuclear fusion process, the suitable nucleiare [2002]
For nuclear fusion process the nuclei with low mass are suitable.
Solar energy is mainly caused due to [2003]
As a result of fusion, enormous amount of heat is liberated which is the main cause of source of solar energy.
A nuclear reaction is given by represents
1^{e0} represents a βdecay..
A sample of radioactive element has a mass of 10gm at an instant t=0. The approximate mass ofthis element in the sample after two mean lives is [2003]
Using the relation for mean life.
Given :
Then from
The mass number of a nucleus is [2003]
In case of hydrogen atom, Mass number = atomic number
The mass of proton is 1.0073 u and that ofneutron is 1.0087 u (u = atomic mass unit). Thebinding energy of is [2003]
Δm = (2 × 1.0074 + 2 × 1.0087 – 4.0015) = 0.0307
E = (Δm) × 931 MeV = 0.0307 × 931 = 28.5 MeV
The volume occupied by an atom is greater thanthe volume of the nucleus by a factor of about[2003]
Hence, V_{atom }= 10^{15 }× V_{nucleus}
If in nuclear fusion process the masses of thefusing nuclei be m1 and m2 and the mass of theresultant nucleus be m3, then [2004]
m_{3} < (m_{1} + m_{2}) (∵ m_{1 }+ m_{2} = m_{3} + E ]
as E = [m_{1} + m_{2} – m_{3}] C^{2}
A nucleus represented by the symbol has
has Z protons and (A – Z) neutrons
Mp denotes the mass of a proton and Mn that ofa neutron. A given nucleus, of binding energyB, contains Z protons and N neutrons. The mass M (N, Z) of the nucleus is given by (c is thevelocity of light) [2004]
Mass defect
Mass of nucleus = Mass of proton + mass of neutron – mass defect
The nuclei of which one of the following pairs of nuclei are isotones? [2005]
Isotones means equal number of neutrons i.e., (A–Z) = 74 –34 = 71 – 31 = 40.
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