Find the maximum energy that a beta particle can have in the following decay 176Lu→176Hf+e+vˉ. Atomic mass of 176Lu is 175.942694amu and that of 176Hf is 175.941420amu

Δm=175.942694−175.941420=0.001274Energy emitted = 0.001274×932=1.182MeV

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Mass-energy equivalence

Find the maximum energy that a beta particle can have in the f

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Find the maximum energy that a beta particle can have in the following decay $^{176} Lu \to^{176} Hf + e + \overset{v}{ˉ} .$ Atomic mass of $^{176} Lu$ is $175.942694 amu$ and that of $^{176} Hf$ is $175.941420 amu$

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Book:

Physics Galaxy Optics and Modern Physics Vol 4 (Ashish Arora)

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Find the maximum energy that a beta particle can have in the following decay

^{176} Lu \to^{176} Hf + e + \overset{v}{ˉ} .

Atomic mass of

^{176} Lu

175.942694 amu

and that of

^{176} Hf

175.941420 amu

8 mins ago

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Text solutionVerified

Δ m = 175.942694 - 175.941420 = 0.001274

Energy emitted =

0.001274 \times 932 = 1.182 MeV

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Practice questions from Physics Galaxy Optics and Modern Physics Vol 4 (Ashish Arora)

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Question Text	Find the maximum energy that a beta particle can have in the following decay $^{176} Lu \to^{176} Hf + e + \overset{v}{ˉ} .$ Atomic mass of $^{176} Lu$ is $175.942694 amu$ and that of $^{176} Hf$ is $175.941420 amu$
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	112