Question

Calculate the de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 keV.

i need the correct and exact answer....​

Answer 1

To Find Given:

Calculate the de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1kV.

Solution:

$\textbf{From the equation: } \: \frac{1}{2 {}^{} } {mv}^{2} = eV$

$e = \text{charge on electron } = 1.6 \times {10}^{ - 19 \: } \text{Coulombs}$

$⇒V = 1000v$

$\textbf{mass of electron = } m = 9.1 \times {10}^{ - 31} kg$

we get,

$⇒1/2 \times 9.1 \times {10}^{ - 31} \times {v}^{2} = 1.6 \times {10}^{ - 19} \times 1000$

$⇒ {v}^{2} = \frac{2 \times 1.6x {10}^{ - 19} \times 1000}{9.1x {10}^{ - 31} }$

$= \frac{3.2 \times {10}^{ - 19} \times 1000}{9.1x {10}^{ - 31} } = 3.5 \times {10}^{4}$

$⇒v = \sqrt{3.5 \times {10}^{4} } = 1.88 \times {10}^{7} \: m/s$

$\text{de-Broglie wavelength of an electron}$

$= λ = \frac{h}{mv}$

$⇒λ = \frac{6.67 \times {10}^{ - 34} }{9.1 \times {10}^{ - 31} \times 1.88 \times {10}^{7} } = 3.87 \times {10}^{ - 11} m$

$\\ \\ \\ \\ \sf \colorbox{lightgreen} {\red★ANSWER ᵇʸɴᴀᴡᴀʙ \: khan}$