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 $=6.63 \times 10^{-34} \mathrm{JS}$
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
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