Advertisements
Advertisements
Question
Why is the detection of neutrinos found very difficult?
Advertisements
Solution
Neutrinos interact very weakly with matter so, they have a very high penetrating power. That’s why the detection of neutrinos is found very difficult.
APPEARS IN
RELATED QUESTIONS
Consider the D−T reaction (deuterium−tritium fusion)
\[\ce{^2_1H + ^3_1H -> ^4_2He + n}\]
Calculate the energy released in MeV in this reaction from the data:
`"m"(""_1^2"H")`= 2.014102 u
`"m"(""_1^3"H")`= 3.016049 u
Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γdecays in the decay scheme shown in Fig. 13.6. You are given that
m (198Au) = 197.968233 u
m (198Hg) =197.966760 u
Write the basic nuclear process underlying β+ and β– decays.
Write the β-decay of tritium in symbolic form.
Write the basic nuclear process of neutron undergoing β-decay.
What is the difference between cathode rays and beta rays? When the two are travelling in space, can you make out which is the cathode ray and which is the beta ray?
In beta decay, an electron (or a positron) is emitted by a nucleus. Does the remaining atom get oppositely charged?
During a negative beta decay,
Ten grams of 57Co kept in an open container beta-decays with a half-life of 270 days. The weight of the material inside the container after 540 days will be very nearly
A free neutron beta-decays to a proton with a half-life of 14 minutes. (a) What is the decay constant? (b) Find the energy liberated in the process.
(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.)
