Advertisements
Advertisements
Question
Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of `""_88^226 "Ra"`.
Given `"m"(""_88^226"Ra")` = 226.02540 u, `"m"(""_86^222 "Rn")` = 222.01750 u,
`"m"(""_86^220 "Rn")`= 220.01137 u, `"m"(""_84^216 "Po")`= 216.00189 u.
Advertisements
Solution
Alpha particle decay of `""_88^226"Ra"` emits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction.
\[\ce{^226_88 Ra -> ^222_86 Ra + ^4_2He}\]
Q-value of
emitted α-particle = (Sum of initial mass − Sum of final mass) c2
Where,
c = Speed of light
It is given that:
`"m"(""_88^226"Ra")` = 226.02540 u
`"m"(""_86^222"Rn")` = 222.01750 u
`"m"(""_2^4"He")` = 4.002603 u
Q-value = [226.02540 − (222.01750 + 4.002603)] u c2
= 0.005297 u c2
But 1 u = 931.5 MeV/c2
∴ Q = 0.005297 × 931.5 ≈ 4.94 MeV
Kinetic energy of the α-particle = `("Mass number after decay"/"Mass number before decay") xx "Q"`
` = 222/226 xx 4.94 = 4.85 " MeV"`
RELATED QUESTIONS
In the study of Geiger-Marsdon experiment on scattering of α particles by a thin foil of gold, draw the trajectory of α-particles in the coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study.
From the relation R = R0 A1/3, where R0 is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A
Two stable isotopes of lithium `""_3^6"Li"` and `""_3^7"Li"` have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.
Boron has two stable isotopes, `""_5^10"B"` and `""_5^11"B"`. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of `""_5^10"B"` and `""_5^11"B"`.
The nucleus `""_10^23"Ne"` decays by `beta^(-)`emission. Write down the β decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
`"m"(""_10^23 "Ne")` = 22.994466 u
`"m"(""_11^23 "Na")` = 22.989770 u.
What do you mean by polar molecules and non-polar molecules? Give ‘one’ example each.
Show that the density of nucleus over a wide range of nuclei is constant-independent of mass number A.
The mass number of a nucleus is
Potassium-40 can decay in three modes. It can decay by β−-emission, B*-emission of electron capture. (a) Write the equations showing the end products. (b) Find the Q-values in each of the three cases. Atomic masses of `""_18^40Ar` , `""_19^40K` and `""_20^40Ca` are 39.9624 u, 39.9640 u and 39.9626 u respectively.
(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)
What is a neutrino?
Atomic mass unit (u) is defined as ________ of the mass of the carbon (12C) atom.
Nuclear species or nuclides are shown by the notation ________ where X is the chemical symbol of the species.
All nuclides with same mass number A are called ______.
Nuclides with same neutron number N but different atomic number Z are called ______.
Before the neutrino hypothesis, the beta decay process was throught to be the transition, `n -> p + vece`. If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range.
James Chadwick, in 1932 studied the emission of neutral radiations when Beryllium nuclei were bombarded with alpha particles. He concluded that emitted radiations were neutrons and not photons. Explain.
What conclusion is drawn from Rutherford’s scattering experiment of α-particles?
What is ‘Pair production’?
Assertion (R): Fusion of hydrogen nuclei into helium nuclei is the source of energy of all stars.
Reason (R): In fusion heavier nuclei split to form lighter nuclei.
