Step 1: Given Charge of alpha particle: q=2e=2×1.6×10−19 Radius of circular path: R=0.83‌cm=0.83×10−2m Strength of magnetic field: B=0.25‌Wb∕m2 Step 2: Formula Used R‌=‌
mv
Bq
λ‌=‌
h
mv
Step 3: Find the de Broglie wavelength Calculate the value of the product of mass and velocity R‌=‌
mv
Bq
⇒mv‌=RBq ⇒mv‌=0.83m×10−2×0.25×2×1.6×10−19 Calculate the de Broglie wavelength by using the formula λ=‌
h
mv
and substituting the values λ‌=‌
6.6×10−34
0.83×10−2×0.25×2×1.6×10−19
‌=‌
6.6×10−34
6.6×10−22
‌=1×10−12 ‌=0.01A Hence, Option (D) is correct. the de Broglie wavelength is 0.01A.