Question

the wavelength of the de- broglie wave associated by an electron accelerate with 100 V is​

Answer 1

According to de-Broglie hypothesis, the wavelength of the wave association with electron is given by:

$\boxed{ \bf{ \lambda = \dfrac{h}{ \sqrt{2meV} } }}$

• h → Plank's Constant $\rm (6.63 \times 10^{-34} \ Js)$
• m → Mass of electron $\rm (9.1 \times 10^{-32} \ kg)$
• e → Charge on electron $\rm (1.6 \times 10^{-19} \ C)$
• V → Potential $\rm (100 \ V)$

By substituting values we get:

$\rm \leadsto \lambda = \dfrac{h}{ \sqrt{2meV} } \\ \\ \rm \leadsto \lambda = \dfrac{6.63 \times 10^{-34} \ Js}{ \sqrt{2 \times 9.1 \times 10^{-32} \ kg \times 1.6 \times 10^{-19} \ C \times V} } \\ \\ \bf \leadsto \lambda = \sqrt{ \dfrac{150}{V} } \: \text{\AA} \\ \\ \rm \leadsto \lambda = \sqrt{ \dfrac{150}{100} } \: \text{\AA} \\ \\ \rm \leadsto \lambda = \sqrt{1.5} \: \text{\AA} \\ \\ \rm \leadsto \lambda = 1.2 \: \text{\AA}$

$\therefore$ de-Broglie wavelength of electron $\sf (\lambda)$ = $\sf 1.2 \ \text{\AA}$