Question

If ${\lambda }_0$ is the de Broglie wavelength for a proton accelerated through a potential difference of 100V, the de Broglie wavelength for $\alpha$ -particle accelerated through the same potential difference is

• A. $2\sqrt{2}{\lambda }_0$
• B. $\frac{{\lambda }_0}{2}$
• C. $\frac{{\lambda }_0}{2\sqrt{2}}$
• D. $\frac{{\lambda }_0}{\sqrt{2}}$

Answer 1

Answer: (C)

Proton is accelerated through a potential difference 100 V
$so,\frac{1}{2}m{v}^2=e\times 100$

$mv=\sqrt{2m_{p}e\times 100}$

${\lambda }_o=\frac{h}{\sqrt{2me\times 100}}$

$Debrogliewavelengthof\alpha -particle$

${\lambda }_{\alpha }=\frac{h}{\sqrt{2m_{\alpha }{v}_{\alpha }\times 100}}$

${\lambda }_{\alpha }=\frac{h}{\sqrt{2\times 4m\times 2e\times 100}}$

$so{\lambda }_{\alpha }=\frac{{\lambda }_0}{2\sqrt{2}}$

So, the answer is option (C).