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 $\left.\lambda \propto \frac{1}{v}\right)$