d
Radius of the circular path of a charge particle in a magnetic field is given by

\(R=\frac{m v}{B q}\)  or  \(m v=R B q\)

Here, \(R=0.83\,\mathrm{cm}=0.83 \times 10^{-2}\, \mathrm{m}\)

\(B=0.25\,\mathrm{Wb}\mathrm{m}^{-2}\)

\(q=2e=2 \times 1.6 \times 10^{-19} \mathrm{C}\)

\(\therefore mv = \) \(\left( {0.83 \times {{10}^{ - 2}}} \right)(0.25)\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)\)

de Broglie wavelength,

\(\lambda  = \frac{h}{{mv}} = \) \(\frac{{6.6 \times {{10}^{ - 34}}}}{{0.83 \times {{10}^{ - 2}} \times 0.25 \times 2 \times 1.6 \times {{10}^{ - 19}}}}\)

\( = 0.01\,\mathop A\limits^o \)