Question

Calculate the de Broglie wavelength of an electron in the first Bohr orbit of a hydrogen atom if the velocity of an electron in the first orbit is \(2.2\times 10^6\text{ ms}^{-1}\). [mass of electron \(= 9.1\times 10^{-31}\) kg, plank's constant (h) \(= 6.626\times 10^{-34}\) J s]

• A. \(3.31\times 10^{-10}\) m
• B. \(3.01\times 10^{-10}\) m
• C. \(3.62\times 10^{-10}\) m
• D. \(2.71\times 10^{-10}\) m

Hint:

Use lambda = h / (m v) with the given velocity.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The de Broglie wavelength of a particle is \(\lambda = \dfrac{h}{mv}\).

Step 2: Substitute.
\[ \lambda = \frac{6.626\times 10^{-34}}{9.1\times 10^{-31}\times 2.2\times 10^{6}} \]
The denominator is \(9.1\times 2.2\times 10^{-25} = 20.02\times 10^{-25} = 2.002\times 10^{-24}\).

Step 3: Calculate.
\[ \lambda = \frac{6.626\times 10^{-34}}{2.002\times 10^{-24}} = 3.31\times 10^{-10}\text{ m} \]

Step 4: Check.
This equals the circumference of the first orbit \(2\pi a_0 = 2\pi\times 0.529\) Å \(\approx 3.32\) Å, as the Bohr condition requires for \(n = 1\).

Final Answer:
The wavelength is \(3.31\times 10^{-10}\) m, option (A). \[ \boxed{3.31\times 10^{-10}\text{ m}} \]