Given:
m16=2.66 x 10^-26 kg
v=3.7 x 10^6 m/s
B= 2.0
These are singly charged hence
q=1.6 x 10^-19 C
The ratio of these two masses is 16 to 18.
Let the mass of m18 be x
m18/m16= x/(2.66 x 10^-26)
18/16= x/(2.66 x 10^-26)
x= [18 x (2.66 x 10^-26)]/16
=2.99 X 10^-26 kg
When they hit a target after traversing a semicircle the distance between them is the difference of their diameter.(∆d)
∆d=2r18-2r16
Where r18 is the radius of the m18 mass
r16 is the radius of the m16 mass.
∆d=2(r18-r16).
The radius of an object transversing in a magnetic field is given by the below formula.
r = mv/qB
m is the mass of the object.
v is the velocity of the object
q is the charge carried by the object
B magnetic field
∆d=2(r18-r16). Substituting the value for r from the above formula.
∆d=2[(m18-m16)v]/qB
m18-m16=
(2.99-2.66)x10^-26=0.33x 10^-26
qB=1.6x10^-19 x2=3.2 x10^-19
Substituting these values in the ∆d formula we get
∆d=
2x3.7x10^6x0.33x10^-26/3.2x10^-19
=2.442 x 10^-20/3.2x 10^-19
=0.76 x10-1m
The separation between the paths while traversing a semicircle is 0.49 m.
What is Magnetic Field?
The region where a particle experiences a magnetic force is known as the magnetic field.
Given data:
The mass of oxygen-16 is, [tex]m = 2.66 \times 10^{-26} \;\rm kg[/tex].
The velocity of single charged particle is, [tex]v = 3.70 \times 10^{6} \;\rm m/s[/tex].
The strength of magnetic field is, B = 2.00 T.
When the particle traverse along a semi-circular path, then the centripetal force will act on the particle, balanced by the magnetic force to maintain its motion.
Therefore,
Fc = Fb
[tex]\dfrac{mv^{2}}{r} = B \times q \times v\\\\r = \dfrac{mv}{Bq}[/tex]
here,
r is the separation (in m) between the paths. And q is the charge of an electron.
Solving as,
[tex]r = \dfrac{2.66 \times 10^{-26}\times 3.70 \times 10^{6}}{2 \times (1.6 \times 10^{-19})}\\\\r = 0.49 \;\rm m[/tex]
Thus, we can conclude that the separation between the paths while traversing a semicircle is 0.49 m.
Learn more about the magnetic force here:
https://brainly.com/question/13791875
