The nucleus of an atom is located at `x=y=z=0`. If the probability of finding an electron in `d_(x^(2)-y^(2))` orbital in a tiny volume around `x=a, y=0, z=0` is `1xx10^(-5)`, what is the probability of finding the electron in the same size volume around `x=0, y=a, z=0` ?

Options

A. `1 xx10^(-5)`
B. `1xx10^(-5)xxa`
C. `-1xx10^(-5)xxa`
D. zero

Explanation

To solve the problem, we need to analyze the probability of finding an electron in a specific orbital and how it relates to the spatial coordinates given.

### Step-by-Step Solution:

1. **Understanding the Orbitals**:
The problem mentions the `d_(x^2 - y^2)` orbital. This orbital has a specific shape and orientation in space. However, it is important to note that the `d_(x^2 - y^2)` orbital is directional, meaning that the probability density of finding an electron varies depending on the coordinates.

2. **Given Probability**:
We are given that the probability of finding the electron in a tiny volume around the point `(x = a, y = 0, z = 0)` is `1 x 10^(-5)`.

3. **Analyzing the Coordinates**:
We need to find the probability of finding the electron in the same size volume around the point `(x = 0, y = a, z = 0)`.

4. **Properties of the Orbital**:
The `d_(x^2 - y^2)` orbital has lobes along the x and y axes. The probability density function for this orbital will yield different values depending on the coordinates due to its directional nature.

5. **Evaluating the Probability**:
For the point `(x = a, y = 0, z = 0)`, the probability is given as `1 x 10^(-5)`. When we evaluate the point `(x = 0, y = a, z = 0)`, we realize that this point is symmetric to the first point with respect to the origin.

6. **Conclusion**:
Since the `d_(x^2 - y^2)` orbital is symmetric about the x and y axes, the probability of finding the electron in the volume around `(x = 0, y = a, z = 0)` will also be the same as that around `(x = a, y = 0, z = 0)`. Thus, the probability remains `1 x 10^(-5)`.

### Final Answer:
The probability of finding the electron in the same size volume around `(x = 0, y = a, z = 0)` is also `1 x 10^(-5)`.

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