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: Find the momentum.
$p = mv = 9.1\times10^{-31}\times 2.2\times10^{6} = 2.002\times10^{-24}$ kg m/s.

Step 2: Divide Planck constant by momentum.
$\lambda = \dfrac{6.626\times10^{-34}}{2.002\times10^{-24}} = 3.31\times10^{-10}$ m.

Step 3: Compare.
The nearby options 3.01, 3.62 and 2.71 (all $\times10^{-10}$) do not match.

Final Answer:
Option (A). \[ \boxed{3.31\times 10^{-10}\text{ m}} \]