Respuesta :
Answer:
Radius = 6.877mm
Step-by-step explanation:
From the question, we have a single charged ion of 7Li with;
Mass(m) = 1.16 × 10^(−26) kg
Potential difference(V) = 250 V
Electron charge of e = 1.6 x 10^(-19)
To solve this question, we'll equate the kinetic energy to potential energy. Thus;
(1/2)(mv²) = eV
Thus;
Velocity (v) =√(2eV/m)
v = √((2 x 1.6 x 10^(-19) x 250)/1.16 × 10^(−26))
v = √(689.655 x 10^(7) = 8.3 x 10⁴ m/s
Since it enters a magnetic field with magnitude B = 0.874 T perpendicular to the path of the ion, thus θ = 90°
Now,we know that;
F = qvBsinθ
Thus F = qvBsin(90)
Sin 90° = 1 ; thus, F =qvB
Also, from Newton's second law, F=ma, also, we know that radial acceleration a = v²/r
So F = mv²/r
Thus equating the 2 forces, we have ;
qvB = mv²/r
So making r the subject, we have;
r = mv/qB = (1.16 × 10^(−26) x 8.3 x 10⁴)/(1.6 x 10^(-19) x 0.874)
= 6.877 x 10^(-3)m = 6.877mm
So, the required radius is [tex]r=6.88\times 10^{-3}m[/tex]
Magnetic field:
The formula for the magnetic field is,
[tex]\frac{1}{2}mu^2=qv\\ u=\frac{2qv}{m}[/tex]
Given that,
Mass=[tex]1.16\times10^{-26} kg[/tex]
From the above formula we have,
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times250}{1.16 \times 10^{-26}} } \\v=8.3\times 10^4 m/s[/tex]
And [tex]r=\frac{mv}{qB}[/tex] then from this formula,
[tex]r=\frac{1.16\times 10^{-26}\times 8.3\times 10^4}{1.6 \times 10^{-19}\times 0.874} \\r=6.88 mm\\or\ r=6.88 \times 10^{-3}m[/tex]
Learn more about the topic Magnetic field:
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