A particle having electric charge $3 \times 10^{-19}$ C and mass $6 \times 10^{-27}$ kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}$ m. The value of $\alpha$ is _________. (Take Planck's constant $= 6.6 \times 10^{-34}$ J.s)
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For calculations involving square roots, simplify the powers of 10 to an even exponent before taking the root to avoid errors.
Step 1: Understanding the Concept:
According to the de Broglie hypothesis, every moving particle has an associated wave whose wavelength is inversely proportional to its momentum.
When a charged particle is accelerated through a potential difference, its kinetic energy increases at the expense of electrical potential energy.
Step 2: Key Formula or Approach:
The de Broglie wavelength is:
\[ \lambda = \frac{h}{p} \]
The kinetic energy $K$ of a particle with charge $q$ accelerated through potential $V$ is:
$K = qV$
Relating kinetic energy to momentum $p$:
$K = \frac{p^2}{2m} \Rightarrow p = \sqrt{2mK} = \sqrt{2mqV}$
Therefore:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \]
Step 3: Detailed Explanation:
Given:
$q = 3 \times 10^{-19}$ C
$m = 6 \times 10^{-27}$ kg
$V = 1.21$ V
$h = 6.6 \times 10^{-34}$ J.s
Calculate the denominator first:
\[ \sqrt{2mqV} = \sqrt{2 \times (6 \times 10^{-27}) \times (3 \times 10^{-19}) \times 1.21} \]
\[ \sqrt{2mqV} = \sqrt{36 \times 1.21 \times 10^{-46}} \]
\[ \sqrt{2mqV} = 6 \times 1.1 \times 10^{-23} = 6.6 \times 10^{-23} \text{ kg}\cdot\text{m/s} \]
Now, calculate the wavelength $\lambda$:
\[ \lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}} = 10^{-11} \text{ m} \]
The question asks for the wavelength in the form $\alpha \times 10^{-12}$ m:
$10^{-11} \text{ m} = 10 \times 10^{-12} \text{ m}$
So, $\alpha = 10$
Step 4: Final Answer:
The value of $\alpha$ is 10.
