The nuclides with odd number of both protons and neutrons are the least stable because, odd number of protons and neutrons results in the presence of two unpaired nucleons.
These unpaired nucleons result in instability. Hence, such nuclides are the least stable.
Probability of success is 3/6=1/2.
`thereforep=1/2andq=1/2`
(i) For exactly 4 successes, the required probability is `""^(7)C_(4)((1)/(2))^(4)((1)/(2))^(3)=35/128`
(ii) For at least 4 successes, the required probability is `""^(7)C_(4)((1)/(4))^(4)((1)/(2))^(3)+""^(7)C_(5)((1)/(2))^(5)((1)/(2))^(2)`
`+""^(7)C_(6)((1)/(2))^(6)((1)/(2))^(1)+""^(7)C_(7)((1)/(2))^(7)`
`=35/128+21/128+7/128+1/128=64/128=1/2`
Nuclides with even number of protons (Z) and even number of neutrons (N) are most stable.
These nuclides tend to form proton-proton and neutron-neutron pairs.
This impart stability to the nucleus.
Beta decay occurs when an unstable nucleus emits a beta particle and energy. A beta particle is either an electron or a positron. An electron is a negatively charged particle,...
Isotopes :
\(_{17}^{38}Cl\) and \(_{17}^{35}Cl\)
Isobars :
\(_6^{14}C\) and \(_{7}^{14}N\)
Isotones :
i. \(_1^3H\) and \(_2^4He\)
ii. \(_{16}^{32}S\) and \(_{15}^{31}P\)
For the given nuclide,
Atomic number, Z = 18,
Mass number, A = 40
Number of protons = Number of electrons = Z = 18
Number of neutrons (N) = A – Z = 40...
Correct Option is : (C) 18, 15, 16
Atomic Number = Number of Protons
Mass Number = Number of Protons + Number of Neutrons
Number of Electron = Atomic Number -(Number of Charges)
Therefore,
Number of...
