Question

Electron coming out from cathode is accelearated towards anode, if the gain in kinetic energy is $400\text{eV}$, what should be the distance between the two plates?
$[$Assume electric field between anode and cathod is $40\text{kN/C}]$.

• A. $1\text{mm}$
• B. $10\text{cm}$
• C. $100\text{mm}$
• D. $1\text{cm}$

Answer 1

The correct option is D $1\text{cm}$


From the given condition, we can say that
gain in $K.E=qV$, and here, $q=e$
$\Rightarrow 400e=eV$
$\therefore V=400\text{V}$
Now, as we know electric field is the negative gradient of potential difference
$\therefore E=\frac{V}{d}$
$\Rightarrow 40\times {10}^3=\frac{400}{d}$
$\Rightarrow d={10}^{-2}\text{m}=1\text{cm}$
Hence, $\left(D\right)$ is the right answer.