Question

The de Broglie wavelength of an electron in the third Bohr orbit of H-atom is

• A. \( 3\pi \times 5.29 \, \text{pm} \)
• B. \( 4\pi \times 52.9 \, \text{pm} \)
• C. \( 6\pi \times 52.9 \, \text{pm} \)
• D. \( 2\pi \times 5.29 \, \text{pm} \)

Hint:

- Bohr's quantization condition: \( mvr = n\frac{h}{2\pi} \). - de Broglie wavelength: \( \lambda = h/p = h/(mv) \). - Combining these leads to \( n\lambda = 2\pi r_n \), meaning the circumference of the orbit is an integer multiple of the de Broglie wavelength. - Radius of n-th Bohr orbit for Hydrogen: \( r_n = n^2 a_0 \), where \( a_0 \approx 52.9 \, \text{pm} \). - So, for the n-th orbit, \( \lambda = \frac{2\pi (n^2 a_0)}{n} = 2\pi n a_0 \).

Answer 1

The Correct Option is C

According to Bohr's quantization condition, the angular momentum of an electron in the n-th orbit is quantized: \[ mvr = \frac{nh}{2\pi} \] where \(m\) is mass of electron, \(v\) is velocity, \(r\) is radius of orbit, \(n\) is principal quantum number, \(h\) is Planck's constant.
The de Broglie wavelength \( \lambda \) is given by \( \lambda = \frac{h}{p} = \frac{h}{mv} \).
From Bohr's condition, \( mv = \frac{nh}{2\pi r} \).
Substitute this into the de Broglie wavelength equation: \[ \lambda = \frac{h}{nh/(2\pi r)} = \frac{h \cdot 2\pi r}{nh} = \frac{2\pi r}{n} \] This means that the circumference of the n-th Bohr orbit is an integral multiple of the de Broglie wavelength: \( n\lambda = 2\pi r_n \).
For the third Bohr orbit of H-atom, \( n=3 \).
The radius of the n-th Bohr orbit of H-atom is \( r_n = a_0 n^2 \), where \( a_0 \) is the Bohr radius.
Given \( a_0 \approx 52.
9 \) pm.
Radius of the third orbit \( r_3 = a_0 (3)^2 = 9a_0 = 9 \times 52.
9 \) pm.
Now, the de Broglie wavelength for \( n=3 \): \[ \lambda = \frac{2\pi r_3}{3} = \frac{2\pi (9a_0)}{3} = 2\pi (3a_0) = 6\pi a_0 \] Substitute \( a_0 = 52.
9 \) pm: \[ \lambda = 6\pi \times 52.
9 \, \text{pm} \] This matches option (3).