A singly ionized atom of Boron [B^+] and a doubly ionized Neon [Ne^++] atom are

Question

A singly ionized atom of Boron [B^+] and a doubly ionized Neon [Ne^++] atom are each accelerated through a potential difference of 5 kV. They enter a 'cloud chamber' where a uniform magnetic field of 4.2 T is directed perpendicular to the plane of the atom's motion. As viewed from above, each atom follows a path that curves to the right. What Is the ratio of the kinetic energy of the Neon atom to that of the Boron? (K_ne/K_B) What is the ratio of the radius of curvature of the Neon to the Boron? (R_ne/R_B) If the atoms lose 1 pJ of kinetic energy for each cm of distance traveled, how far does each atom travel before coming to rest?

Explanation / Answer

charge on Boron atom = e = 1.609e-19 C

charge on Neon atom = 2e

mass of Bron ion = 10

mass of Neon ion =20

electric potential V = 5 kV

magnetic fild B = 4.2 T

KE gained by Boron = 5.0e+3 * 1.6e-19 = 8.0e-16 J

KE of Neon             = 5.0e+3*2*1.6e-19 = 16.0 e-16 J

Kne /Kb = 16/8 =2

b) raidus of curvature of charged particle ina magnetic field

R = mv/Bq

mas of Neon Mn = 33.18 e-27 kg

speed of Neon vn = sqrt(2*16.0e-16/33.18e-27) = 3.11e+5 m/s

mass of Boron Mb = 18.26e-27 kg

speed of Boron vb = sqrt(2*8.0e-16/18.26e-27) = 2.96 e+5 m/s

Rne = 33.18e-27*3.11e+5/(4.2*2*1.6e-19 ) = 7.675 e-3 m

Rb   = 18.26e-27 *2.96e+5/(4.2*1.6e-19) = 8.04 e-3 m

Rne/ Rb = 7.675/8.04 = 0.954

c) KE of Neon = 16.0e-16 J

energy for 1 cm = 1pJ

distance traveld by Ne = 16.0e-16/1.0e-12 = 1.6e-3 cm

KE of Boron = 8.0e-16 J

distance travelled by Boron = 8.0e-16 /1.0e-12 = 8.0e-4 cm

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