Question

The energy of separation of an electron in a Hydrogen atom in excited state is 3.4 eV. The de-Broglie wavelength (in Å) associated with the above electron is, if radius of first orbit of H-atom is 0.53Å

• A. 3.33
• B. 6.66
• C. 13.31
• D. 4.2

Answer 1

Answer: (B)

6.66

Answer 2

The problem asks for the de-Broglie wavelength associated with an electron in a hydrogen atom. We are given the energy of separation of the electron in an excited state, which is 3.4 eV. This means the energy of the electron in that excited state is $E_n = -3.4$ eV.

1. Determine the principal quantum number (n): The energy of an electron in the nth orbit of a hydrogen atom is given by: $E_n = -\frac{13.6}{n^2} \text{ eV}$

Given $E_n = -3.4 \text{ eV}$: $-3.4 = -\frac{13.6}{n^2}$ $n^2 = \frac{13.6}{3.4}$ $n^2 = 4$ $n = 2$ (Since n must be a positive integer, the electron is in the second excited state, i.e., n=2 orbit).

2. Relate de-Broglie wavelength to Bohr's orbit: According to Bohr's quantization condition, for a stable orbit, the circumference of the orbit ($2\pi r_n$) must be an integral multiple of the de-Broglie wavelength ($\lambda$): $2\pi r_n = n\lambda$ Therefore, $\lambda = \frac{2\pi r_n}{n}$

3. Use the formula for the radius of the nth orbit: The radius of the nth orbit in a hydrogen atom is given by: $r_n = n^2 a_0$ where $a_0$ is the Bohr radius (radius of the first orbit) = 0.53 Å.

4. Substitute $r_n$ into the de-Broglie wavelength equation: $\lambda = \frac{2\pi (n^2 a_0)}{n}$ $\lambda = 2\pi n a_0$

5. Calculate the de-Broglie wavelength: Substitute $n=2$ and $a_0 = 0.53 \text{ Å}$: $\lambda = 2\pi (2) (0.53 \text{ Å})$ $\lambda = 4\pi (0.53 \text{ Å})$ Using $\pi \approx 3.14159$: $\lambda = 4 \times 3.14159 \times 0.53 \text{ Å}$ $\lambda = 12.56636 \times 0.53 \text{ Å}$ $\lambda = 6.6591708 \text{ Å}$

Rounding to two decimal places, $\lambda \approx 6.66 \text{ Å}$.