Question

If the velocity of the electron in Bohr's first orbit is \(2.19\times 10^6 \text{Ms}^{-1}\), Calculate the de Broglie wavelength associated with it.
\([h = 6.626\times 10^{-34} \text{J s}\) & Mass of electrons \(= 9.10938\times 10^{-31}\) kg \(]\)

• A. \(332\) pm
• B. \(313\) pm
• C. \(342\) pm
• D. \(323\) pm

Hint:

Use lambda = h/(m v) with the given speed and convert metres to picometres.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Every moving particle has an associated wavelength given by de Broglie. For an electron of mass \(m\) and speed \(v\), the wavelength is \(\lambda = h/(mv)\).

Step 2: Key Formula or Approach
\[ \lambda = \frac{h}{m v} \]

Step 3: Detailed Explanation
Put in the values:
\[ \lambda = \frac{6.626 \times 10^{-34}}{9.10938 \times 10^{-31} \times 2.19 \times 10^{6}} \]
Denominator: \(9.10938 \times 2.19 = 19.95\), so it is \(1.995 \times 10^{-24}\).
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.995 \times 10^{-24}} = 3.32 \times 10^{-10} \text{ m} \]
Since 1 pm = \(10^{-12}\) m, \(\lambda = 332\) pm. This equals the circumference of the first Bohr orbit (2 pi times 52.9 pm), as expected for one full wave. The other options are nearby numbers that do not follow from the data.

Final Answer:
The de Broglie wavelength is 332 pm, option (A). \[ \boxed{332 \text{ pm}} \]