Let's solve the problem step-by-step: **Given:** We are to find the probability of finding a 2s electron of hydrogen atom at: (i) \( r = a_0 \) (Bohr radius) (ii) \( r = 2a_0 \) where \( a_0 \) is the Bohr radius. --- ## 1. **Write the Radial Wave Function for 2s Electron** The normalized radial wave function for the 2s electron of hydrogen atom is: \[ R_{2s}(r) = \frac{1}{4\sqrt{2\pi}} \left( \frac{1}{a_0^{3/2}} \right) \left(2 - \frac{r}{a_0}\right) e^{-r/2a_0} \] --- ## 2. **Probability Density** The probability of finding the electron between \( r \) and \( r+dr \) is: \[ P(r)dr = |R_{2s}(r)|^2 \cdot 4\pi r^2 dr \] So, the **probability density** at distance \( r \) is: \[ P(r) = |R_{2s}(r)|^2 \cdot 4\pi r^2 \] --- ## 3. **Calculate at \( r = a_0 \) and \( r = 2a_0 \)** ### **(i) At \( r = a_0 \):** #### **Step 1: Calculate \( R_{2s}(a_0) \)** \[ R_{2s}(a_0) = \frac{1}{4\sqrt{2\pi}} \frac{1}{a_0^{3/2}} \left(2 - \frac{a_0}{a_0}\right) e^{-a_0/2a_0} = \frac{1}{4\sqrt{2\pi}} \frac{1}{a_0^{3/2}} (2 - 1) e^{-1/2} \] \[ = \frac{1}{4\sqrt{2\pi}} \frac{1}{a_0^{3/2}} (1) e^{-1/2} \] #### **Step 2: Calculate \( |R_{2s}(a_0)|^2 \)** \[ |R_{2s}(a_0)|^2 = \left[ \frac{1}{4\sqrt{2\pi}} \frac{1}{a_0^{3/2}} e^{-1/2} \right]^2 = \frac{1}{16 \times 2\pi} \frac{1}{a_0^3} e^{-1} = \frac{1}{32\pi a_0^3} e^{-1} \] #### **Step 3: Multiply by \( 4\pi r^2 \) at \( r = a_0 \)** \[ P(a_0) = |R_{2s}(a_0)|^2 \cdot 4\pi a_0^2 = \frac{1}{32\pi a_0^3} e^{-1} \cdot 4\pi a_0^2 \] \[ = \frac{4\pi a_0^2}{32\pi a_0^3} e^{-1} = \frac{4}{32} \frac{1}{a_0} e^{-1} \] \[ = \frac{1}{8a_0} e^{-1} \] #### **Final Answer for (i):** \[ \boxed{ P(a_0) = \frac{1}{8a_0} e^{-1} } \] --- ### **(ii) At \( r = 2a_0 \):** #### **Step 1: Calculate \( R_{2s}(2a_0) \)** \[ R_{2s}(2a_0) = \frac{1}{4\sqrt{2\pi}} \frac{1}{a_0^{3/2}} \left(2 - \frac{2a_0}{a_0}\right) e^{-2a_0/2a_0} \] \[ = \frac{1}{4\sqrt{2\pi}} \frac{1}{a_0^{3/2}} (2 - 2) e^{-1} = 0 \] #### **Step 2: Calculate \( |R_{2s}(2a_0)|^2 \)** \[ |R_{2s}(2a_0)|^2 = 0^2 = 0 \] #### **Step 3: Probability at \( r = 2a_0 \)** \[ P(2a_0) = |R_{2s}(2a_0)|^2 \cdot 4\pi (2a_0)^2 = 0 \] #### **Final Answer for (ii):** \[ \boxed{ P(2a_0) = 0 } \] --- ## **Summary Table** | \( r \) | Probability Density \( P(r) \) | |-----------------|----------------------------------------| | \( a_0 \) | \( \dfrac{1}{8a_0} e^{-1} \) | | \( 2a_0 \) | \( 0 \) | --- ## **Explanation** - The probability density at a point is calculated using the radial wave function squared, multiplied by \( 4\pi r^2 \). - For the hydrogen 2s orbital, the radial function has a node at \( r = 2a_0 \), so the probability density is zero at that radius. - At \( r = a_0 \), substitute into the formula to get the probability density. --- ## **Final Answers** **(i) At \( r = a_0 \):** \[ \boxed{P(a_0) = \frac{1}{8a_0} e^{-1}} \] **(ii) At \( r = 2a_0 \):** \[ \boxed{P(2a_0) = 0} \]