Question

$$^{67}\mathrm{Ho}(165)$$ is a stable isotope. Which decay mode is possible for $$^{1}H^{3}$$?

A. $\alpha$-emission
B. $\beta$-emission
C. Positron emission
D. $\gamma$-emission

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Answer

The question asks about the possible decay mode of the isotope $^{3}_{1}\mathrm{H}$.

The correct answer is C: पॉजिट्रान उत्सर्जन (Positron Emission).

Reasoning:

  • पॉजिट्रान उत्सर्जन (Positron Emission) occurs when a proton in the nucleus is converted into a neutron, releasing a positron ($(\beta^+)$) and a neutrino ($(\nu)$).
  • The isotope $^{3}_{1}\mathrm{H}$ (also known as Tritium) has an excess of protons compared to the neutrons, making it likely to undergo positron emission to reach a more stable state.

So, the most probable decay mode for $^{3}_{1}\mathrm{H}$ is पॉजिट्रान उत्सर्जन (Positron Emission).

Translated to English:

The question asks about the possible decay mode of the isotope $^{3}_{1}\mathrm{H}$.

The correct answer is C: Positron Emission.

Reasoning:

  • Positron Emission occurs when a proton in the nucleus is converted into a neutron, releasing a positron ($(\beta^+)$) and a neutrino ($(\nu)$).
  • The isotope $^{3}_{1}\mathrm{H}$ (also known as Tritium) has an excess of protons compared to the neutrons, making it likely to undergo positron emission to reach a more stable state.

So, the most probable decay mode for $^{3}_{1}\mathrm{H}$ is Positron Emission.


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