The minimum uncertainty in de Broglie wavelength of an electron accelerated from rest by a potential

Question

The minimum uncertainty in de Broglie wavelength of an electron accelerated from rest by a potential difference of 6.0V, if the uncertainty in measuring the position is 1/π nm, is

a) 6.25 Å (b) 6.0 Å (c) 0.625 Å (d) 0.3125 Å

Accepted Answer

$\Delta x.\Delta y \geq \frac{\lambda^{2}}{4\pi}\text{ and }\lambda = \sqrt{\frac{150}{6}} = 5\overset{o}{A}$

$\therefore\Delta{\lambda\frac{\lambda^{2}}{4\pi.\Delta x}\frac{\left( 5 \times 10^{- 10} \right)^{2}}{4\pi \times \left( \frac{1}{\pi} \times 10^{- 9} \right)}^{- 11}\text{ m}}_{\min}$

Question

Solution

More Atomic Structure Questions