MCQ
In given graph if $V_{02}>V_{01}$ then __________ .
- A$\lambda_1=\sqrt{\lambda_2}$
- B$\lambda_1<\lambda_2$
- C$\lambda_1=\lambda_2$
- D$\lambda_1>\lambda_2$
✓
Answer
D. $\lambda_1>\lambda_2$
$\rightarrow V _0=\left(\frac{h}{e}\right) v-\frac{\phi_0}{e}$ and
From graph $\left( V _0\right)_2>\left( V _0\right)_1$
$\begin{array}{l}
\therefore\left(\frac{h}{e}\right) v_2-\frac{\phi_0}{c}>\left(\frac{h}{e}\right) v_1-\frac{\phi_0}{e} \\
\therefore v_2>v_1 \quad\left[\because V_0=\frac{h v}{c}-\frac{\phi_0}{e}\right] \\
\therefore \lambda_2<\lambda_1 \quad\left[\because \lambda \propto \frac{1}{v}\right] \\
\therefore \lambda_1>\lambda_2
\end{array}$
$\rightarrow V _0=\left(\frac{h}{e}\right) v-\frac{\phi_0}{e}$ and
From graph $\left( V _0\right)_2>\left( V _0\right)_1$
$\begin{array}{l}
\therefore\left(\frac{h}{e}\right) v_2-\frac{\phi_0}{c}>\left(\frac{h}{e}\right) v_1-\frac{\phi_0}{e} \\
\therefore v_2>v_1 \quad\left[\because V_0=\frac{h v}{c}-\frac{\phi_0}{e}\right] \\
\therefore \lambda_2<\lambda_1 \quad\left[\because \lambda \propto \frac{1}{v}\right] \\
\therefore \lambda_1>\lambda_2
\end{array}$
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