Question

Example 2.14 An electron is moving with a kinetic energy
of 4.55 x 10-25 J. Calculate its de-Broglie wavelength.
1 2
1
mu​

Answer 1

ANSWER:

• The de-Broglie wavelength of the electron = 7.28 x 10⁻³² m.

GIVEN:

• An electron is moving with a kinetic energy of 4.55 x 10⁻²⁵ J.

TO FIND:

• The de-Broglie wavelength of the electron.

EXPLANATION:

$\boxed{ \boldsymbol{ \large{ \orange{\lambda =\dfrac{h}{\sqrt{2m\ K.E}}}}}}$

$\implies\sf h = 6.626 \times 10^{-34}\ Js$

$\implies\sf m= 9.1 \times 10^{-31}\ kg$

$\implies\sf K.E=4.55 \times 10^{-25}\ J$

$\sf\leadsto\lambda =\dfrac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 4.55 \times 10^{-25}}}$

$\sf\leadsto\lambda =\dfrac{6.626 \times 10^{-34}}{\sqrt{18.2 \times 10^{-6} \times 4.55}}$

$\sf\leadsto\lambda =\dfrac{6.626 \times 10^{-34}}{\sqrt{82.81 \times 10^{-6}}}$

$\sf\leadsto\lambda =\dfrac{6.626 \times 10^{-34}}{9.1 \times 10^{-3}}$

$\sf\leadsto\lambda =0.728 \times 10^{-31}$

$\sf\leadsto\lambda =7.28 \times 10^{-32} \ m$

Hence the de-Broglie wavelength of the electron = 7.28 x 10⁻³² m.