Electrons in an experiment are accelerated by a voltage of 36 kV. If the voltage is increased to 144 kV, then the de-broglie wavelength associated with the electron would

decreased by 2-times

The correct answer is Option (2) → decreased by 2-times

The momentum of an electron (P) that has been accelerated by a voltage V is -

$P=\sqrt{2meV}$

$⇒P_1=\sqrt{2×9.11×10^{-21}×1.6×10^{-19}×36×10^3}$

$≃3.63×10^{-24}kg\,m^2/s^2$

$⇒P_2=\sqrt{2×9.11×10^{-31}×1.6×10^{-19}×144×10^3}$

$≃7.26×10^{-24}kg\,m^2/s^2$

Now,

$λ=\frac{h}{P}$ [De-Broglie]

$\frac{λ_1}{λ_2}=\frac{\frac{h}{P_1}}{\frac{h}{P_2}}=\frac{P_2}{P_1}$

$⇒\frac{λ_1}{λ_2}=\frac{7.26×10^{-24}}{3.63×10^{-24}}=2$