Question
Consider the situation of the previous problem. Consider the fastest electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate.
✓
Answer
Here electric field of metal plate $=\text{E}=\frac{\text{p}}{\text{E}_0}$
$=\frac{1\times10^{-19}}{8.85\times10^{-12}}=113\text{v/m}$
accl. de $=\phi=\frac{\text{qE}}{\text{m}}$
$=\frac{1.6\times10^{-19}\times113}{9.1\times10^{-31}}=19.87\times10^{12}$
$\text{t}=\frac{\sqrt{2\text{y}}}{\text{a}}=\frac{\sqrt{2\times20\times10^{-2}}}{19.87\times10^{-31}}=1.41\times10^{-7}\text{sec}$
K.E. $=\frac{\text{hc}}{\lambda}-\text{w}=1.2\text{ev}$
$= 1.2 × 1.6 × 10 - 19J [$because in previous problem i.e. in problem $31 : KE = 1.2ev]$
$\therefore\text{v}=\frac{\sqrt{2\text{KE}}}{\text{m}}$
$=\frac{\sqrt{2\times1.2\times1.6\times10^{-19}}}{4.1\times10^{-31}}=0.665\times10^{-6}$
$\therefore$ Horizontal displacement $= Vt × t$
$= 0.655 \times 10^{-6} \times 1.4 \times 10^{-7} = 0.092m = 9.2cm$.
$=\frac{1\times10^{-19}}{8.85\times10^{-12}}=113\text{v/m}$
accl. de $=\phi=\frac{\text{qE}}{\text{m}}$
$=\frac{1.6\times10^{-19}\times113}{9.1\times10^{-31}}=19.87\times10^{12}$
$\text{t}=\frac{\sqrt{2\text{y}}}{\text{a}}=\frac{\sqrt{2\times20\times10^{-2}}}{19.87\times10^{-31}}=1.41\times10^{-7}\text{sec}$
K.E. $=\frac{\text{hc}}{\lambda}-\text{w}=1.2\text{ev}$
$= 1.2 × 1.6 × 10 - 19J [$because in previous problem i.e. in problem $31 : KE = 1.2ev]$
$\therefore\text{v}=\frac{\sqrt{2\text{KE}}}{\text{m}}$
$=\frac{\sqrt{2\times1.2\times1.6\times10^{-19}}}{4.1\times10^{-31}}=0.665\times10^{-6}$
$\therefore$ Horizontal displacement $= Vt × t$
$= 0.655 \times 10^{-6} \times 1.4 \times 10^{-7} = 0.092m = 9.2cm$.
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