The wavelength of electrons accelerated from rest through a poten… - Vidyadip

MCQ

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 electrons $=9.1 \times 10^{-31}\, \mathrm{~kg}$

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

Planck's constant $=6.63 \times 10^{-34\,} \mathrm{Js}$

A
$9$
B
$4$
C
$5$
✓
$6$

✓

Answer

Correct option: D.

$6$

d
De-broglie-wave length of electron:

$\lambda_{c}=\frac{h}{\sqrt{2 m(\mathrm{KE})}}$

$\because \mathrm{e}^{-}\,{\text {is accelerated }}{\text { from rest }}{\Rightarrow \mathrm{KE}=\mathrm{q} \times \mathrm{V}}$

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

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

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

$=6.16 \times 10^{-12} \,\mathrm{~m}$

Nearest integer $=6$

OR

$=\frac{12.3}{\sqrt{V}} \,\mathring A$

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

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