$\Delta p =m v=\sqrt{2 m K}$
$=\sqrt{2 m e V}$
where, $K=$ kinetic energy, $m=$ mass of electron, $v=$ velocity of electron and $e=$ charge of electron.
Wavelength $\lambda$ associated with these moving electrons is
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m e V}}$
where, $h$ is the Planck's constant.
Fringe width $\beta$ shown by electrons in Young's double slit experiment, is
$\beta=\frac{\lambda D}{ d }$
where, $D$ is the distance of screen from fringes and $d$ is the separation of slits. $\therefore \quad \beta=\frac{h D}{d \sqrt{2 m e V}}$ $\Rightarrow \beta \propto \frac{1}{\sqrt{V}}$, keeping other parameters constant.
So, ratio of initial and final fringe widths is
$\Rightarrow \frac{\beta_{i}}{\beta_{f}}=\frac{\sqrt{V_{f}}}{\sqrt{V_{i}}}=\frac{\sqrt{2 V_{i}}}{\sqrt{V_{i}}}$
$\Rightarrow \beta_{f}=\frac{\beta_{i}}{\sqrt{2}}=0.7 \beta_{i}$
$\Rightarrow \left.\beta_{f}=0.7(\omega) \quad \text { (given, } \beta_{i}=\omega\right)$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| $List-I$ | $List-II$ |
| ($P$) If $n=2$ and $\alpha=180^{\circ}$, then all the possible values of $\theta_0$ will be | ($1$) $30^{\circ}$ and $0^{\circ}$ |
| ($Q$) If $n=\sqrt{3}$ and $\alpha=180^{\circ}$, then all the possible values of $\theta_0$ will be | ($2$) $60^{\circ}$ and $0^{\circ}$ |
| ($R$) If $n=\sqrt{3}$ and $\alpha=180^{\circ}$, then all the possible values of $\phi_0$ will be | ($3$) $45^{\circ}$ and $0^{\circ}$ |
| ($S$) If $n=\sqrt{2}$ and $\theta_0=45^{\circ}$, then all the possible values of $\alpha$ will be | ($4$) $150^{\circ}$ |
| ($5$) $0^{\circ}$ |
