The nucleus `._48^115 Cd` after two successive `beta^-` decays will give.

Options

A. `._(46)Pa^(115)`
B. `._(49) ln^(114)`
C. `._(50)Sn^(113)`
D. `._(50)Sn^(115)`

Answer

To solve the problem of determining the nucleus that results from two successive beta-minus (β^-) decays of the cadmium nucleus \( _{48}^{115}Cd \), we can follow these steps:

### Step 1: Understand Beta Minus Decay
In beta-minus decay, a neutron in the nucleus is converted into a proton, and a beta particle (electron) is emitted. This process increases the atomic number (Z) by 1 while the mass number (A) remains unchanged.

### Step 2: Write the Initial Nucleus
The initial nucleus is cadmium, represented as:
\[ _{48}^{115}Cd \]
where:
- Atomic number (Z) = 48
- Mass number (A) = 115

### Step 3: First Beta Minus Decay
After the first beta decay, the atomic number increases by 1:
\[ Z = 48 + 1 = 49 \]
The mass number remains the same:
\[ A = 115 \]
Thus, the new nucleus after the first decay is:
\[ _{49}^{115}In \]
(where In is indium).

### Step 4: Second Beta Minus Decay
Now, we perform a second beta decay on the indium nucleus:
Again, the atomic number increases by 1:
\[ Z = 49 + 1 = 50 \]
The mass number remains the same:
\[ A = 115 \]
Thus, the new nucleus after the second decay is:
\[ _{50}^{115}Sn \]
(where Sn is tin).

### Final Answer
After two successive beta-minus decays, the nucleus \( _{48}^{115}Cd \) transforms into:
\[ _{50}^{115}Sn \]