Question

An electron, a proton, a deuteron and an $\alpha -$ particle are accelerated by the same potential $V$. If ${\lambda }_e,{\lambda }_p,{\lambda }_d$ and ${\lambda }_{\alpha }$ represents the de-Broglie wavelengths of electron, proton, deuteron and $\alpha -$ particle respectively. Which of the following statement(s) is/are correct.

• A. ${\lambda }_p=\frac{0.202}{\sqrt{V}}\mathring{A}$
• B. ${\lambda }_{\alpha }=\frac{0.101}{\sqrt{V}}\mathring{A}$
• C. ${\lambda }_d=\frac{0.280}{\sqrt{V}}\mathring{A}$
• D. ${\lambda }_e>{\lambda }_p>{\lambda }_d>{\lambda }_{\alpha }$

Answer 1

Answer: (B), (D)

${\lambda }_e>{\lambda }_p>{\lambda }_d>{\lambda }_{\alpha }$

If a charged particle with charge $\left(q\right)$ and mass $\left(m\right)$ is accelerated through a potential difference $V$. Then the de-Broglie wavelength associated with the charged particle is given by,

$\lambda =\frac{h}{\sqrt{2mqV}}$

$\because$ The de-Broglie wavelength associated with an electron is given by,

$\Rightarrow {\lambda }_e=\frac{12.27}{\sqrt{V}}\mathring{A}$

For proton :

$q=e:m=m_p=1836m_e$

$\Rightarrow {\lambda }_p=\frac{12.27}{\sqrt{1836}\sqrt{V}}\mathring{A}$

$\Rightarrow {\lambda }_p=\frac{0.286}{\sqrt{V}}\mathring{A}$

Hence, option $\left(A\right)$ is wrong.

For deuteron ;

$q=e:m=2m_p$

$\Rightarrow {\lambda }_d=\frac{{\lambda }_p}{\sqrt{2}\sqrt{V}}=\frac{0.286}{\sqrt{2}\sqrt{V}}$

${\lambda }_d=\frac{0.202}{\sqrt{V}}\mathring{A}$

Hence option $\left(C\right)$ is wrong.

For $\alpha -$ particle ;

$q=2e;m=4m_p$

$\Rightarrow {\lambda }_{\alpha }=\frac{{\lambda }_p}{\sqrt{8}\sqrt{V}}=\frac{0.286}{\sqrt{8}\sqrt{V}}$

$\Rightarrow {\lambda }_{\alpha }=\frac{0.101}{\sqrt{V}}\mathring{A}$

Hence, option $\left(B\right)$ is correct.

On comparing all the four values of de-Broglie wavelength, we get

${\lambda }_e>{\lambda }_p>{\lambda }_d>{\lambda }_{\alpha }$

Hence, option $\left(D\right)$ is correct.

Therefore, options $\text{(B) and (D)}$ are correct.