AIPMT 2012 Physics and Chemistry Paper

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.