Atomic weight of single charge Li is 6.06 amu and its energy is 400 eV . It enter normally to a uniform magnetic field of 0.8 Tesla, then radius of circular path is:

Question

Atomic weight of single charge Li is 6.06 amu and its energy is 400 eV . It enter normally to a uniform magnetic field of 0.8 Tesla, then radius of circular path is: (A)  8.35 cm  (B)  9.23 cm  (C)  10.5 cm  (D)  11.25 cm

Tardigrade Admin · Accepted Answer

Radius of circular path r=(m upsilon /Bq)=(√2mE/Bq) =(√2× 6.06× 1.6× 10-27× 400× 1.6× 10-19/1.6× 10-19× 0.8) =8.35cm

Q. Atomic weight of single charge $L i$ is $6.06 am u$ and its energy is $400 e V$ . It enter normally to a uniform magnetic field of $0.8$ Tesla, then radius of circular path is:

$8.35 c m$

$9.23 c m$

$10.5 c m$

$11.25 c m$

Radius of circular path $r = \frac{m υ}{Bq} = \frac{2 m E}{Bq}$ $= \frac{2 \times 6.06 \times 1.6 \times 10 ^{- 27} \times 400 \times 1.6 \times 10 ^{- 19}}{1.6 \times 10 ^{- 19} \times 0.8}$ $= 8.35 c m$

Q. Atomic weight of single charge Li is 6.06amu and its energy is 400eV . It enter normally to a uniform magnetic field of 0.8 Tesla, then radius of circular path is:

8.35cm

9.23cm

10.5cm

11.25cm