Explanation (Complete Step-by-Step)

When a charged particle is accelerated through potential V:

\[ \text{Kinetic Energy} = qV \]

\[ \frac{1}{2}mv^2 = qV \]

Momentum:

\[ p = mv \]

From kinetic energy relation:

\[ v = \sqrt{\frac{2qV}{m}} \]

\[ p = m \sqrt{\frac{2qV}{m}} \]

\[ p = \sqrt{2mqV} \]

De Broglie wavelength:

\[ \lambda = \frac{h}{p} \]

\[ \lambda = \frac{h}{\sqrt{2mqV}} \]

Substitute values:

\[ h = 6.6 \times 10^{-34} \]

\[ m = 6 \times 10^{-27} \]

\[ q = 3 \times 10^{-19} \]

\[ V = 1.21 \]

\[ \lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 6 \times 10^{-27} \times 3 \times 10^{-19} \times 1.21}} \]

First calculate inside root:

\[ 2 \times 6 \times 3 \times 1.21 = 43.56 \]

\[ = 43.56 \times 10^{-46} \]

\[ \sqrt{43.56 \times 10^{-46}} = 6.6 \times 10^{-23} \]

\[ \lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}} \]

\[ \lambda = 10^{-11} \text{ m} \]

\[ \lambda = 10 \times 10^{-12} \text{ m} \]

Therefore, \( \alpha = 10 \)

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