Question

For two particles A and B, graphs are plotted for $\sqrt{V}$ against the de broglie wavelengths $\lambda$, where $V$ is the potential on the particles. Which of the following relation is correct about the mass of the particles?

• A. $m_A=m_B$
• B. $m_A<m_B$
• C. $m_A>m_B$
• D. $m_A\ge m_B$

Answer 1

The correct option is C $m_A>m_B$

de Broglie relation,
$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2m(K.E.)}}=\frac{h}{\sqrt{2m\left(qV\right)}}$
where, $\lambda$, $v$, $h$, $q$, $V$ and $m$ are wavelength, velocity, Planck's constant, charge, potential and mass of a particle respectively.

For the same value of $\lambda$, $V$ and $m$ are related as:
$\lambda \propto \frac{1}{m}$
So, at any wavelength $\left(\lambda \right)$, $V_B>V_A$.
Similarly, $V\propto \frac{1}{m}$
then, $m_A>m_B$