Question

(D) size of cathode
49. The minimum uncertainty in the position of electron moving with a velocity of 3 x 107 m/s will be:
(A) 0.0388​

Answer 1

Given:

Electron moving with a velocity of 3 × 10⁷ m/s.

To find:

Minimum uncertainty in position of electron.

Calculation:

Applying Heisenberg's Uncertainty Principle:

$\therefore \: \Delta x \times \Delta p \geqslant \dfrac{h}{4\pi}$

$= > \: \Delta x \times (m\Delta v)\geqslant \dfrac{h}{4\pi}$

$= > \: \Delta x \geqslant \dfrac{h}{4\pi \times m\Delta v}$

$= > \: \Delta x \geqslant \dfrac{6.63 \times {10}^{ - 34} }{4\pi \times (9.1 \times {10}^{ - 31} \times 3 \times {10}^{7}) }$

$= > \: \Delta x \geqslant \dfrac{2.21 \times {10}^{ - 34} }{4\pi \times (9.1 \times {10}^{ - 31} \times {10}^{7}) }$

$= > \: \Delta x \geqslant \dfrac{2.21 \times {10}^{ - 34} }{4\pi \times (9.1 \times {10}^{ - 24} ) }$

$= > \: \Delta x \geqslant 0.019 \times {10}^{ - 10} \: m$

$= > \: \Delta x \geqslant 0.019 \: {A}^{ \circ}$

So, final answer is:

$\boxed{ \sf{ \red{ \large{ \: \Delta x \geqslant 0.019 \: {A}^{ \circ} }}}}$