Question

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

Answer 1

Calculating  the de-Broglie wavelength of an electron :
*****************************************************

From the equation:
1/2 mv²=ev------------1
e= charge on electron=1.6 x 10⁻¹⁹ Coulombs
V=1000v
mass of electron=m=9.1 x10⁻³¹ kg

Substituting these values in equation 1, we get 
1/2 x9.1 x10⁻³¹ x v² = 1.6 x 10⁻¹⁹ x1000
v²=2x1.6 x 10⁻¹⁹ x1000/9.1 x10⁻³¹
v²=3.2x10⁻¹⁹x1000/9.1 x10⁻³¹
v²=3.5x101⁴
v=√3.5x10⁴=1.88 x 10⁷ m/s

de-Broglie wavelength of an electron = λ=h/mv
where h= 6.67 x 10 ⁻³⁴ J-s
λ= 6.67 x 10 ⁻³⁴ /9.1 x10⁻³¹ x1.88 x 10⁷
=3.87 x10⁻¹¹ m

∴de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 keV is 3.87 x10⁻¹¹ m

Answer 2

Answer:

4.13×10^-11

Explanation:

HOPE IT HELPS YOU!

ALL THE BEST