An electron is accelerated through a potential difference of 144 V. The de-broglie wavelength associated would be approximately:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.1 nm
The kinetic energy is also related,
$K.E.=\frac{1}{2}mv^2$
$ev=\frac{1}{2}mv^2⇒v=\sqrt{\frac{2ev}{m}}$
$v=\sqrt{\frac{2ev}{m}}=\sqrt{\frac{2×1.6×10^{-19}×144}{9.11×10^{-31}}}$
$≃7.12×10^6$
Using the Momentum formula,
$P=mv=(7.12×10^6)×(9.11×10^{-31})=6.48×10^{-24}$
Wavelength, $λ=\frac{h}{P}=\frac{6.63×10^{-34}}{6.48×10^{-24}}$
$=0.1nm$