Question

The mass of an electron is \(9.1\times10^{-31}\ \text{kg}\). If its K.E. is \(3.0\times10^{-25}\ \text{J}\), its wavelength is (approximately):

• A. $250\text{ nm}$
• B. $990\text{ nm}$
• C. $400\text{ nm}$
• D. $850\text{ nm}$

Hint:

When performing rapid calculations in competitive exams, simplify the power of ten first:
Here, the denominator power of ten is $\sqrt{10^{-56}} = 10^{-28}$.
Dividing $10^{-34}$ by $10^{-28}$ gives $10^{-6}\text{ m}$ (which is in the range of hundreds of nanometers). This instantly helps narrow down the choices.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to calculate the de Broglie wavelength ($\lambda$) of an electron given its mass ($m$) and kinetic energy ($\text{K.E.}$).


Step 2: Key Formula or Approach:
The de Broglie wavelength formula in terms of kinetic energy is:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(\text{K.E.})}} \]
Where:
- $h$ is Planck's constant $\approx 6.626 \times 10^{-34}\text{ J s}$ (or $\text{kg m}^2\text{ s}^{-1}$)
- $m = 9.1 \times 10^{-31}\text{ kg}$
- $\text{K.E.} = 3.0 \times 10^{-25}\text{ J}$


Step 3: Detailed Explanation:
Let us calculate the term inside the square root first:
\[ 2m(\text{K.E.}) = 2 \times (9.1 \times 10^{-31}\text{ kg}) \times (3.0 \times 10^{-25}\text{ J}) \]
\[ 2m(\text{K.E.}) = 54.6 \times 10^{-56}\text{ kg}^2\text{ m}^2\text{ s}^{-2} \]
Now, take the square root of this value:
\[ \sqrt{2m(\text{K.E.})} = \sqrt{54.6} \times 10^{-28}\text{ kg m s}^{-1} \]
Since $\sqrt{54.6} \approx 7.39$:
\[ \sqrt{2m(\text{K.E.})} \approx 7.39 \times 10^{-28}\text{ kg m s}^{-1} \]
Now, let us calculate the de Broglie wavelength ($\lambda$):
\[ \lambda = \frac{6.626 \times 10^{-34}}{7.39 \times 10^{-28}} \]
\[ \lambda \approx 0.897 \times 10^{-6}\text{ m} = 897\text{ nm} \approx 850\text{ nm} \]


Step 4: Final Answer:
The correct option is (D).