Question

The two lines $\text{A and B}$ , shown in figure are the graphs of the de-Broglie wavelength $\lambda$ as a function of $\frac{1}{\sqrt{V}}$ $(V$ is the accelerating potential) for two particles having the same charge. Identify the correct option about the mass of the particles-

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

Answer 1

The correct option is B $m_A>m_B$

From the de-Broglie wavelength expression,

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

$\therefore \lambda \propto \frac{1}{\sqrt{V}}$

$\Rightarrow \text{slope}=\frac{\lambda }{\left(\frac{1}{\sqrt{V}}\right)}$

Since, the graph of $\lambda$ vs $\left(\frac{1}{\sqrt{V}}\right)$ is a straight line.

$\Rightarrow \text{slope}=\left(\frac{h}{\sqrt{2mq}}\right)$

Hence, lesser the slope more will be the mass,

From graph , $A$ has less slope than $B$

Therefore, $m_A>m_B$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.