Which equation represents alpha decay?

${ }_{95}^{241} Am \rightarrow{ }_{93}^{237} Np+{ }_2^4 He$

${ }_9^{18} F \rightarrow{ }_8^{18} O +{ }_1^0 e$

${ }_6^{14} C \rightarrow{ }_7^{14} N+{ }_{-1}^0 e$

${ }_{66}^{152} D y \rightarrow{ }_{66}^{152} D y+y

Alpha decay involves the emission of an alpha particle ( 2 4 ​ He ) from a nucleus.
The mass number decreases by 4, and the atomic number decreases by 2 during alpha decay.
The equation 95 241 ​ A m → 93 237 ​ Np + 2 4 ​ He correctly represents alpha decay.
Therefore, the answer is 95 241 ​ A m → 93 237 ​ Np + 2 4 ​ He .

Explanation

Understanding Alpha Decay Alpha decay is a type of radioactive decay where an atomic nucleus emits an alpha particle and transforms into a different atomic nucleus, with a decrease in mass number by 4 and atomic number by 2. An alpha particle is composed of 2 protons and 2 neutrons, which is the same as a helium nucleus ( 2 4 ​ He ). We need to identify the equation that correctly represents this process.

Analyzing the Equations Let's examine each of the given equations:

95 241 ​ A m → 93 237 ​ Np + 2 4 ​ He : In this equation, Americium-241 decays into Neptunium-237 and an alpha particle ( 2 4 ​ He ). The mass number decreases from 241 to 237 (a change of 4), and the atomic number decreases from 95 to 93 (a change of 2). This matches the criteria for alpha decay.

9 18 ​ F → 8 18 ​ O + 1 0 ​ e : In this equation, Fluorine-18 decays into Oxygen-18 and a positron ( 1 0 ​ e ). The mass number remains the same (18), and the atomic number decreases from 9 to 8. This represents positron emission, not alpha decay.

6 14 ​ C → 7 14 ​ N + − 1 0 ​ e : In this equation, Carbon-14 decays into Nitrogen-14 and an electron ( − 1 0 ​ e ). The mass number remains the same (14), and the atomic number increases from 6 to 7. This represents beta decay, not alpha decay.

66 152 ​ Dy → 66 152 ​ Dy + y : This equation shows Dysprosium-152 transforming into itself and some particle 'y'. The mass number and atomic number remain the same, and the emitted particle is not specified as an alpha particle. This does not represent alpha decay.

Identifying the Correct Equation Based on the analysis, the first equation, 95 241 ​ A m → 93 237 ​ Np + 2 4 ​ He , correctly represents alpha decay because it shows the emission of an alpha particle ( 2 4 ​ He ) from the nucleus, with the mass number decreasing by 4 and the atomic number decreasing by 2.


Examples
Alpha decay is used in smoke detectors. Americium-241, an alpha emitter, is used to ionize air within the detector. When smoke particles enter the detector, they disrupt the ionization process, reducing the current. This reduction triggers the alarm. Understanding alpha decay helps in designing and improving such safety devices, ensuring they are effective in detecting smoke and preventing fire hazards. The process relies on the predictable decay of the radioactive material and the properties of alpha particles.

Answered by GinnyAnswer | 2025-07-07

The equation that represents alpha decay is 95 241 ​ A m → 93 237 ​ Np + 2 4 ​ He because it demonstrates the emission of an alpha particle with a decrease in mass number by 4 and atomic number by 2. Other equations either exemplify different decay processes or show no decay at all.
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Answered by Anonymous | 2025-07-27