Question

A proton and an electron are accelerated by the same potential difference. Let${\lambda }_e$ and ${\lambda }_p$ denote the de Broglie wavelengths of the electron and the proton respectively.

• A. ${\lambda }_e={\lambda }_p$
• B. ${\lambda }_e<{\lambda }_p$
• C. ${\lambda }_e>{\lambda }_p$
• D. The relation between ${\lambda }_{e}and{\lambda }_p$depends on the accelerating potential difference

Answer 1

The correct option is C

${\lambda }_e>{\lambda }_p$

Given – a proton and an electron are accelerated by same potential difference Let that be v. Then their de Broglie wavelength is ${\lambda }_e$ and ${\lambda }_p$

$\begin{array}{cc}{\lambda }_e=\frac{h}{p_e}& {\lambda }_p=\frac{h}{p_p}\\ {\lambda }_e=\frac{h}{m_e{v}_e}& {\lambda }_p=\frac{h}{m_p{v}_P}\end{array}$

So as we see here that the wavelength depends on 2 variables.

We cant compare as such.

As$m_e<m_p$ yes

But we don’t know the relation between$V_{e}and{V}_p$

Or do we

There is an extra piece of information

The V-potential difference for accelerating both particles is same

So ………………

So if both had 0 velocity initially, then they were accelerated by same potential to gain velocity $V_{e}and{V}_p$

Lets find a relation

Lets conserve energy

$\Delta$ K.E.=$\frac{1}{2}m{v}_e^2$
$\Delta$ P.E.=qv
$\Rightarrow \frac{1}{2}{m}_e{v}_e^2$=qv
$\frac{1}{2}{m}_e{v}_e^2=m_{e}qv$
P=$m_e{v}_e=\sqrt{2m_{e}qv}$

similarlyP=$m_e{v}_e=\sqrt{2m_{p}qv}$

Substituting in de Broglie’s wavelength

${\lambda }_e$ = $\frac{h}{\sqrt{2m_{e}qV}}\mid {\lambda }_p=\frac{h}{\sqrt{2m_{p}qV}}$

q and V is same for both

only one variable

$m_e<m_p$

$\Rightarrow$ ${\lambda }_e>{\lambda }_p$

Option c is correct