An electron moving with de Broglie wavelength $3.77Å$ describes a circle in a magnetic field of strength $2\times {10}^{-2} \mathrm{T}$ .The diameter of the circle is nearly

For the electron with de Broglie wavelength $\lambda$,

$p=mv=\frac{h}{\lambda } \Rightarrow v=\frac{h}{m\lambda }$.

Radius of path of electron moving with speed v in magnetic field B,

$r=\frac{mv}{eB}=\frac{m}{eB}\left(\frac{h}{m\lambda }\right)=\frac{h}{eB\lambda }$.

Diameter of the circle,

$D=2r=\frac{2h}{eB\lambda }=\frac{2\times \left(6.63\times {10}^{-34}\right)}{\left(1.6\times {10}^{-19}\right)\times \left(2\times {10}^{-2}\right)\times \left(3.77\times {10}^{-10}\right)}=1.1\times {10}^{-3} \mathrm{m}=1.1 \mathrm{mm}$