Question

De-Broglie wavelength of an electron orbiting in the \(n = 2\) state of hydrogen atom is close to (Given Bohr radius = 0.052 nm):

• A. \( 0.67 \text{ nm} \)
• B. \( 1.67 \text{ nm} \)
• C. \( 2.67 \text{ nm} \)
• D. \( 0.067 \text{ nm} \)

Hint:

Remember Bohr's quantization condition \( 2 \pi r = n \lambda \), relating the circumference of the electron's orbit to its de Broglie wavelength. Also, recall the formula for the radius of the \(n^{th}\) Bohr orbit \( r_n = n^2 a_0 \).

Answer 1

The Correct Option is B

Step 1: Bohr's Quantization Condition
According to Bohr's quantization condition, the angular momentum of an electron in the \(n^{th}\) orbit is quantized:
\( L = mvr = n \frac{h}{2\pi} \),
where \(m\) is the mass of the electron, \(v\) is the velocity, \(r\) is the radius, and \(h\) is Planck's constant.

Step 2: de Broglie Wavelength
The de Broglie wavelength of the electron is given by:
\( \lambda = \frac{h}{p} = \frac{h}{mv} \),
where \(p\) is the momentum.
From Bohr's condition, \( mv = \frac{nh}{2\pi r} \).
Substituting this into the de Broglie wavelength equation, we get:
\( \lambda = \frac{h}{\frac{nh}{2\pi r}} = \frac{2\pi r}{n} \).

Step 3: For \(n = 2\)
For the \(n = 2\) state, the wavelength is:
\( \lambda = \frac{2\pi r_2}{2} = \pi r_2 \).
The radius of the \(n^{th}\) Bohr orbit is given by:
\( r_n = n^2 a_0 \), where \(a_0\) is the Bohr radius (\(0.052\) nm).
For \(n = 2\), the radius is:
\( r_2 = 2^2 a_0 = 4 \times 0.052 \text{ nm} = 0.208 \text{ nm} \).

Step 4: Calculate the de Broglie Wavelength
Now, we calculate the de Broglie wavelength for \(n = 2\):
\( \lambda = \pi r_2 = \pi \times 0.208 \text{ nm} \approx 3.14 \times 0.208 \text{ nm} \approx 0.653 \text{ nm} \).

Conclusion
There seems to be a discrepancy with the provided options. However, the closest option to 0.653 nm is 0.67 nm.