Home
Class 12
PHYSICS
The nucleus .^(23)Ne decays by beta-emi...

The nucleus `.^(23)Ne` decays by `beta`-emission into the nucleus `.^(23)Na`. Write down the `beta`-decay equation and determine the maximum kinetic energy of the electrons emitted. Given,`(m(._(11)^(23)Ne) =22.994466 am u` and `m (._(11)^(23)Na =22.989770 am u`. Ignore the mass of antineutrino `(bar(v))`.

Text Solution

Verified by Experts

` "" _(10) ^(23) Ne to "" _(11) ^(23) Na + vece +vecv+ Q`
For ` beta ^(-) -` decay, `Q = [ M(c) - M(y) ]C ^(2)`
` = [ 22.994466-22. 979770] 931.5 MeV`
`= 0.004696 x 931.5 MeV = 4. 37 MeV`
Promotional Banner

Topper's Solved these Questions

  • NUCLEI

    NARAYNA|Exercise EVALUATE YOURSELF-1|9 Videos

  • NUCLEI

    NARAYNA|Exercise EVALUATE YOURSELF-2|9 Videos

  • NUCLEAR PHYSICS

    NARAYNA|Exercise LEVEL-II-(H.W)|9 Videos

  • RAY OPTICS AND OPTICAL INSTRAUMENTS

    NARAYNA|Exercise EXERCISE- 4 One or more than one correct answer type|13 Videos

Similar Questions

Explore conceptually related problems

The nucleus .^(23)Ne deacays by beta -emission into the nucleus .^(23)Na . Write down the beta -decay equation and determine the maximum kinetic energy of the electrons emitted. Given, (m(._(11)^(23)Ne) =22.994466 am u and m (._(11)^(23)Na =22.989770 am u . Ignore the mass of antineuttino (bar(v)) .

.^(23)Ne decays to .^(23)Na by negative beta emission. What is the maximum kinetic enerfy of the emitter electron ?

Complete the decay reaction ._10Ne^(23) to?+._-1e^0+? Also, find the maximum KE of electrons emitted during this decay. Given mass of ._10Ne^(23)=22.994465 u . mass of ._11Na^(23)=22.989768u .

In beta decay, ""_(10)Ne^(23) is converted into ""_(11)Na^(23) . Calculate the maximum kinetic energy of the electrons emitted, given atomic masses of ""_(10)Ne^(23) and ""_(11)Na^(23) arc 22.994466 u and 22.989770 u, respectively.

The nucleus of ._(90)^(230)Th is unstable against alpha- decay with a half-life of 7.6xx10^(3) years. Write down the equation of the decay and estimate the kinetic energy of the emitted alpha- particle from the following data: m(._(90)^(230)Th) =230.033131 amu, m(._(88)^(226)Ra) =226.025406 amu and m(._(2)^(4)He) =4.002603 amu.

A nucleus ""_(10)Ne^(23) undergoes beta - decay and becomes ""_(11)Na^(23) . Calculate the maximum kinetic energy of electrons emitted assuming that the daughter nucleus and antineutrino carry negligible kinetic energy. Given mass of ""_(10)Ne^(23) = 22.994466 u and mass of ""_(11)Na^(23) = 22.989770 m

Gold ._(79)^(198)Au undergoes beta^(-) decay to an excited state of ._(80)^(198)Hg . If the excited state decays by emission of a gamma -photon with energy 0.412 MeV , the maximum kinetic energy of the electron emitted in the decay is (This maximum occurs when the antineutrino has negligble energy. The recoil enregy of the ._(80)^(198)Hg nucleus can be ignored. The masses of the neutral atoms in their ground states are 197.968255 u for ._(79)^(198)Hg ).

The radionuclide ._6C^(11) decays according to ._6C^(11)to ._5B^(11) +e^(+) +v : half life =20.3min. The maximum energy of the emitted positron is 0.960 MeV . Given the mass values m(._6C^(11))=11.011434u, m(._6B^(11))=11.009305u Calculate Q and compare it with maximum energy of positron emitted.