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Let's work through this problem step by step.

The formula for the radial probability density (RPD) in the 1s orbital is given by:

\[
RPD(1s) = \frac{4}{a_0^3} r^2 e^{-2r/a_0} dr
\]

We are asked to calculate the radial probability of finding the electron in the 1s orbital at a radius \( r = 2.50 \, a_0 \), where \( a_0 \) is the Bohr radius. We are also provided the value of the Bohr radius:

\[
a_0 = 5.29 \times 10^{-11} \, \text{m}
\]

### Steps:

1. **Substitute \( r = 2.50 \, a_0 \) into the RPD formula**:
   The equation becomes:

   \[
   RPD(1s) = \frac{4}{a_0^3} (2.50 \, a_0)^2 e^{-2(2.50 \, a_0)/a_0} dr
   \]

2. **Simplify the equation**:
   - The \( r^2 \) term becomes \( (2.50)^2 a_0^2 \).
   - The exponent simplifies to \( e^{-5} \).

   So the equation becomes:

   \[
   RPD(1s) = \frac{4}{a_0^3} \cdot (6.25 a_0^2) \cdot e^{-5} dr
   \]

3. **Simplify further**:
   Cancel out \( a_0^2 \):

   \[
   RPD(1s) = \frac{4 \times 6.25}{a_0} e^{-5} dr
   \]

   \[
   RPD(1s) = \frac{25}{a_0} e^{-5} dr
   \]

4. **Insert the value of \( a_0 \)**:
   Now, substitute \( a_0 = 5.29 \times 10^{-11} \, \text{m} \):

   \[
   RPD(1s) = \frac{25}{5.29 \times 10^{-11}} e^{-5} dr
   \]

5. **Calculate the values**:
   - \( e^{-5} \approx 0.0067379 \)
   - \( \frac{25}{5.29 \times 10^{-11}} \approx 4.7274 \times 10^{11} \)

   Therefore:

   \[
   RPD(1s) = 4.7274 \times 10^{11} \times 0.0067379 \, dr
   \]

6. **Final calculation**:

   \[
   RPD(1s) \approx 3.184 \times 10^9 \, dr
   \]

### Expressing the answer to four decimal places:

The radial probability density is approximately:

\[
RPD(1s) \approx 3.1840 \times 10^9 \, dr
\]

Would you like me to explain any step in further detail or perform additional calculations?

### 5 Related Questions:
1. How would the radial probability change for an electron in the 2s orbital?
2. What is the significance of the Bohr radius in atomic physics?
3. How does the probability density function vary with increasing values of \( r \)?
4. What is the physical meaning of the exponential decay term in the RPD?
5. How does the radial distribution function differ from the probability density?

### Tip:
Always make sure to double-check the units and numerical values when working with constants like the Bohr radius to ensure accuracy in your final results.

What is the radial probability of finding an electron in the 1s orbital at a radius of 2.50 a₀? Note that the dr term will be 5.29 × 10⁻¹¹ (the Bohr Radius). Express your answer to four decimal places (X.XXXX).

Calculate the radial probability of finding an electron in the 1s orbital at a radius of 2.50 Bohr radii. This problem involves quantum mechanics concepts, specifically the radial probability density function and exponential decay. The Bohr radius is used as a constant in the equation.

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