The mass of an electron is $9.1 \times 10^{-31} \mathrm{~kg}$. If its K.E. is $3.0 \times 10^{-25} \mathrm{~J}$,
Question.
The mass of an electron is $9.1 \times 10^{-31} \mathrm{~kg}$. If its K.E. is $3.0 \times 10^{-25} \mathrm{~J}$, calculate its wavelength.
The mass of an electron is $9.1 \times 10^{-31} \mathrm{~kg}$. If its K.E. is $3.0 \times 10^{-25} \mathrm{~J}$, calculate its wavelength.
Solution:
From de Broglie’s equation,
$\lambda=\frac{\mathrm{h}}{m v}$
Given,
Kinetic energy (K.E) of the electron $=3.0 \times 10^{-25} \mathrm{~J}$
Since $\mathrm{K} . \mathrm{E}=\frac{1}{2} m v^{2}$
$\therefore$ Velocity $(v)=\sqrt{\frac{2 \mathrm{~K} \cdot \mathrm{E}}{m}}$
$=\sqrt{\frac{2\left(3.0 \times 10^{-25} \mathrm{~J}\right)}{9.10939 \times 10^{-31} \mathrm{~kg}}}$
$=\sqrt{6.5866 \times 10^{4}}$
$v=811.579 \mathrm{~ms}^{-1}$
Substituting the value in the expression of $\lambda$ :
$\lambda=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)\left(811.579 \mathrm{~ms}^{-1}\right)}$
$\lambda=8.9625 \times 10^{-7} \mathrm{~m}$
Hence, the wavelength of the electron is $8.9625 \times 10^{-7} \mathrm{~m}$.
From de Broglie’s equation,
$\lambda=\frac{\mathrm{h}}{m v}$
Given,
Kinetic energy (K.E) of the electron $=3.0 \times 10^{-25} \mathrm{~J}$
Since $\mathrm{K} . \mathrm{E}=\frac{1}{2} m v^{2}$
$\therefore$ Velocity $(v)=\sqrt{\frac{2 \mathrm{~K} \cdot \mathrm{E}}{m}}$
$=\sqrt{\frac{2\left(3.0 \times 10^{-25} \mathrm{~J}\right)}{9.10939 \times 10^{-31} \mathrm{~kg}}}$
$=\sqrt{6.5866 \times 10^{4}}$
$v=811.579 \mathrm{~ms}^{-1}$
Substituting the value in the expression of $\lambda$ :
$\lambda=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)\left(811.579 \mathrm{~ms}^{-1}\right)}$
$\lambda=8.9625 \times 10^{-7} \mathrm{~m}$
Hence, the wavelength of the electron is $8.9625 \times 10^{-7} \mathrm{~m}$.