Question

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

Answer 1

Answer:

The required de-Broglie wavelength is 3.89 x10⁻¹¹ m

Explanation:

Mass of the electron, m = 9.1 × 10⁻³¹ kg

charge of the electron, e = 1.6 × 10⁻¹⁹ C

Kinetic energy = 1 keV = 1000 eV

 $\sf \dfrac{1}{2}mv^2=1000 eV \\\\ \sf \dfrac{1}{2} \times 9.1 \times 10^{-31} \times v^2=1000 \times 1.6 \times 10^{-19} \\\\ \sf 4.55 \times 10^{-31} \times v^2=1.6 \times 10^{-16} \\\\ \sf v^2=\dfrac{1.6 \times 10^{-16}}{4.55 \times 10^{-31} } \\\\ \sf v^2=0.35 \times 10^{15} \\\\ \sf v^2=3.5 \times 10^{14} \\\\ \sf v=\sqrt{3.5 \times 10^{14}} \\\\ \longrightarrow \sf v=1.87 \times 10^7 \ m/s$

Therefore, the velocity of the electron is 1.87 × 10⁷ m/s

The de-Broglie wavelength is given by,

$\longmapsto \sf \lambda =\dfrac{h}{mv}$

where

h is the Planck's constant (h = 6.626 × 10⁻³⁴ J-s)

Substituting the values,

$\sf \lambda =\dfrac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 1.87 \times 10^7} \\\\ \sf \lambda =0.389 \times 10^{-10} \\\\ \sf \lambda =3.89 \times 10^{-11} \ m$

The de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 keV is 3.89 x10⁻¹¹ m

Answer 2

Answer:

Energy acquired by the electron (as kinetic energy) after being aacelerated by a potential difference of 1 kV.

(i.e. 1000 volts) = 1000 e V = 1000 × 1.602 × 10 − 19 J

= 1.602 × 10 − 16 J ( 1 e V = 1.602 × 10 − 19 J )

=1000eV=1000×1.602×10-19J

=1.602×10-16J(1eV=1.602×10-19J)

i.e., Kinetic energy,

1 2 m v 2 = 1.602 × 10 − 16 J or 1 2 × 9.1 × 10 − 31 v 2 12mv2

=1.602×10-16Jor12×9.1×10-31v2 or

v 2 = 3.521 × 10 14 or

v = 1.88 × 10 7 m s − 1

v2=3.521×1014orv=1.88×107ms-1

∴ λ = h m v

= 6.626 × 10 − 34 k g m 2 s − 1 ( 9.1 × 10 − 31 k g ) × ( 1.88 × 10 7 m s − 1 )

= 3.87 × 10 − 1m

Explanation:

Your telegram?