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3 years ago

Tritium 3 1H decays to 3 2He by beta emission. Find the energy released in the process. Answer in units of keV.

Chemistry

1 answer:

-Dominant- [34]3 years ago

7 0

Answer:

The energy released in the decay process = 18.63 keV

Explanation:

To solve this question, we have to calculate the binding energy of each isotope and then take the difference.

The mass of Tritium = 3.016049 amu.

So,the binding energy of Tritium = 3.016049 *931.494 MeV

= 2809.43155 MeV.

The mass of Helium 3 = 3.016029 amu.

So, the binding energy of Helium 3 = 3.016029 * 931.494 MeV

= 2809.41292 MeV.

The difference between the binding energy of Tritium and the binding energy of Helium is: 32809.43155 - 2809.412 = 0.01863 MeV

1 MeV = 1000keV.

Thus, 0.01863 MeV = 0.01863*1000keV = 18.63 keV.

So, the energy released in the decay process = 18.63 keV.

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RSB [31]

Answer:

$V=0.0310L=3.10mL$

Explanation:

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In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

$n_{acid}=n_{KOH}$

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$n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol$

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3 years ago

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natima [27]

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