Question

In the following diagram if $V_2>V_1$ then

• A. ${\lambda }_1=\sqrt{{\lambda }_2}$
• B. ${\lambda }_1<{\lambda }_2$
• C. ${\lambda }_1={\lambda }_2$
• D. ${\lambda }_1>{\lambda }_2$

Answer 1

Answer: (D)

$\because V_0=\left(\frac{h}{e}\right)v-\left(\frac{W_0}{e}\right).$
From the graph $V_2>V_1$
$\Rightarrow \frac{h{v}_2}{e}-\frac{W_0}{e}>\frac{h{v}_1}{e}-\frac{W_0}{e}$
$\Rightarrow v_2>v_2$
$\Rightarrow {\lambda }_1>{\lambda }_2$ (as $\lambda \propto \frac{1}{v})$