Question

If proton and a - particle are accelerated by the same potential $V$ . Then, the ratio of their wavelengths $\lambda _{\text{P}}:\lambda _{\alpha }$ would be

• A. $2\sqrt{2}:1$
• B. $1 \text{:} 2 \sqrt{2}$
• C. $2:1$
• D. $1:2$

Answer 1

Answer: (A)

$2\sqrt{2}:1$

Answer 2

KE of proton and alpha-particle shall be $1 eV$ and $2 eV$ respectively. Now, momentum, $p =\sqrt{2 mKE }$ $\therefore$ The momentum of proton is $p _{ P }=\sqrt{2 m _{p} \times eV }$ $\therefore$ The momentum of $\alpha$ -particle whose mass $m _{\alpha}=4 m _{ P }$ is $p _{\alpha}=\sqrt{8 m _{p} \times(2 eV )}=\sqrt{16 m _{p} eV }$ $\therefore$ Now, $\lambda=\frac{h}{p} \quad$ [de-Broglie relation] $\Rightarrow \frac{\lambda_{ p }}{\lambda_{\alpha}}=\frac{ p _{\alpha}}{ p _{p}}=\frac{\sqrt{16 m _{p} eV }}{\sqrt{2 m _{p} eV }}=2 \sqrt{2}: 1$