The first orbital of H is represented by: `psi=(1)/(sqrtr)((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 of finding the probability of locating the electron at a distance \( r \) from the nucleus in the hydrogen atom, we can follow these steps:

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

### Step 2: Calculate the Probability Density
The probability density \( P \) is given by the square of the wave function:
\[ P = |\psi|^2 = \psi \cdot \psi^* = \left(\frac{1}{\sqrt{r}} \left(\frac{1}{a_0}\right)^{3/2} e^{-r/a_0}\right)^2 \]
This simplifies to:
\[ P = \frac{1}{r} \left(\frac{1}{a_0}\right)^3 e^{-2r/a_0} \]

### Step 3: Determine the Volume Element
In spherical coordinates, the volume element \( dV \) at a distance \( r \) is given by:
\[ dV = 4\pi r^2 dr \]

### Step 4: Calculate the Probability of Finding the Electron
The probability \( dP \) of finding the electron in the shell between \( r \) and \( r + dr \) is given by:
\[ dP = P \cdot dV = \left(\frac{1}{r} \left(\frac{1}{a_0}\right)^3 e^{-2r/a_0}\right) \cdot (4\pi r^2 dr) \]
This simplifies to:
\[ dP = 4\pi \left(\frac{1}{a_0}\right)^3 e^{-2r/a_0} r dr \]

### Step 5: Final Expression
Thus, the probability of finding the electron at a distance \( r \) from the nucleus in the region \( dV \) is:
\[ dP = \frac{4\pi}{a_0^3} r e^{-2r/a_0} dr \]