Question

What is the de-Broglie wavelength of the $\alpha -$particle accelerated through a potential difference V.

• A. $\frac{0.287}{\sqrt{V}}$Å
• B. $\frac{12.27}{\sqrt{V}}$Å
• C. $\frac{0.101}{\sqrt{V}}$Å
• D. $\frac{0.202}{\sqrt{V}}$Å

Answer 1

Answer: (C)

$\frac{0.101}{\sqrt{V}}$Å

Answer 2

$\lambda = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2m_{\alpha}Q_{\alpha}V}}$

On putting $Q_{\alpha} = 2 \times 1.6 \times 10^{- 19}C$

$m_{\alpha} = 4m_{p} = 4 \times 1.67 \times 10^{- 27}kg$ ⇒ $\lambda = \frac{0.101}{\sqrt{V}}Å$