Question

Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is $1.6 \times 10^6 \,m \,s^{-1}$, the de Broglie wavelength associated with this electron is

• A. 455 pm
• B. 222 pm
• C. 333 pm
• D. 755 pm

Answer 1

Answer: (A)

According to de Broglie equation, $\lambda=\frac{h}{m v}$
where, $ h =$ Planck's constant $=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} $
$m =$ mass of electron $=9.1 \times 10^{-31} \mathrm{~kg} $
$v=$ velocity $=1.6 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$
$\therefore \lambda =\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 1.6 \times 10^{6}}$
$=0.455 \times 10^{-9} \mathrm{~m} $
$=455 \times 10^{-12} \mathrm{~m}=455\, \mathrm{pm}$