The first orbital of H is represented by: `psi=(1)/(sqrtpi)((1)/(a_(0)))^(3//2)e^(-r//a_(0))`, where `a_(0)` is Bohr's radius. The probability of finding the electron at a distance r, from the nucleus in the region dV is :

Options

A. `psi^(2)dr`
B. `int psi^(2)4pir^(2)dv`
C. `psi^(2)4pir^(2)dr`
D. `int psidv`

Answer

To solve the problem, we need to find the probability of finding an electron at a distance \( r \) from the nucleus in the hydrogen atom, given the wave function \( \psi \).

### Step-by-Step Solution:

1. **Identify the Wave Function**:
The wave function for the first orbital of hydrogen is given as:
\[ \psi = \frac{1}{\sqrt{\pi}} \left(\frac{1}{a_0}\right)^{3/2} e^{-r/a_0} \]
where \( a_0 \) is the Bohr radius.

2. **Calculate \( \psi^2 \)**:
The probability density is given by the square of the wave function:
\[ \psi^2 = \left(\frac{1}{\sqrt{\pi}} \left(\frac{1}{a_0}\right)^{3/2} e^{-r/a_0}\right)^2 = \frac{1}{\pi} \left(\frac{1}{a_0}\right)^3 e^{-2r/a_0} \]

3. **Set Up the Probability Element**:
The probability of finding the electron in a spherical shell of thickness \( dr \) at a distance \( r \) from the nucleus is given by:
\[ dP = \psi^2 \cdot dV \]
where \( dV \) is the volume element in spherical coordinates:
\[ dV = 4\pi r^2 dr \]
Therefore,
\[ dP = \psi^2 \cdot 4\pi r^2 dr \]

4. **Substitute \( \psi^2 \) into the Probability Element**:
Substitute \( \psi^2 \) into the equation for \( dP \):
\[ dP = \left(\frac{1}{\pi} \left(\frac{1}{a_0}\right)^3 e^{-2r/a_0}\right) \cdot 4\pi r^2 dr \]

5. **Simplify the Expression**:
Simplifying the expression gives:
\[ dP = \frac{4}{a_0^3} e^{-2r/a_0} r^2 dr \]

6. **Conclusion**:
The probability of finding the electron at a distance \( r \) from the nucleus in the region \( dV \) is:
\[ dP = \frac{4}{a_0^3} e^{-2r/a_0} r^2 dr \]

### Final Answer:
The probability of finding the electron at a distance \( r \) from the nucleus in the region \( dV \) is:
\[ dP = \frac{4}{a_0^3} e^{-2r/a_0} r^2 dr \]