The wavelength of electrons accelerated from rest through
Question:

The wavelength of electrons accelerated from rest through a potential difference of $40 \mathrm{kV}$ is $\mathrm{x} \times 10^{-12} \mathrm{~m}$. The value of $\mathrm{x}$ is (Nearest integer)

Given : Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$

Charge on an electron $=1.6 \times 10^{-19} \mathrm{C}$

Planck’s constant $=6.63 \times 10^{-34} \mathrm{JS}$

Solution:

De-broglie-wave length of electron:

$\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}\left\{\begin{array}{l}\because \mathrm{e}^{-} \text {is accelerated } \\ \text { from rest } \\ \Rightarrow \mathrm{KE}=\mathrm{q} \times \mathrm{V}\end{array}\right.$

$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}$

$=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.6 \times 10^{-19} \times 9.1 \times 10^{-31} \times 40 \times 10^{3}}}$

$=0.614 \times 10^{-11} \mathrm{~m}$

$=6.14 \times 10^{-12} \mathrm{~m}$

Nearest integer = 6

OR

$\lambda=\frac{12.3}{\sqrt{\mathrm{V}}} \AA$

$=\frac{12.3}{200}=6.15 \times 10^{-12} \mathrm{~m}$

Ans. is 6