Question

Ions having equal charges but masses of M and 2M are accelerated through the same potential difference and then enter a uniform magnetic field perpendicular to their path. If the heavier ions follow a circular arc of radius R, what is the radius of the arc followed by the lighter?

Answer 1

Answer:

\frac{R}{^{\sqrt{2}}}

Explanation:

q1 = q , q2 = q

m1 = M, m2 = 2M

Potential difference is V

Let v1 be the speed of first particle and v2 be the speed of second particle.

Energy due to potential difference gives the kinetic energy to the particle.

1/2 m x v^2 = e V

v^2 = 2 e V / m = 2e V / m ... (1)

$v}=\sqrt{\frac{2eV}{m}}$

Now radius is given by

r = m v / B q

$r=\frac{\sqrt{2emV}}{Bq}$ by eqution (1)

For first particle

$r_{1}=\frac{\sqrt{2eMV}}{Bq}$   .... (2)

For second particle

$r_{2}=\frac{\sqrt{4eMV}}{Bq}$

$R=\frac{\sqrt{4eMV}}{Bq}$    .... (3)

Divide equation (2) by equation (3), we get

$r_{1}=\frac{R}{^{\sqrt{2}}}$

So

r1 / r2 =

Answer 2

The radius of the arc followed by the lighter is 0.5R.

Kinetic energy of the charges

The kinetic energy of the charges is calculated as follows;

$K.E = \frac{1}{2} mv^2\\\\eV = \frac{1}{2} mv^2\\\\v = \sqrt{\frac{2eV}{m} }$

Force experienced by the charges

$F_c = F_B\\\\\frac{mv^2}{r} = qvB\\\\r = \frac{mv}{qB} \\\\r = \frac{m\sqrt{2eV/m} }{qB}$

Radius of the first particle

$r_1 = \frac{m_1\sqrt{2eV/m_1} }{qB}$

Radius of the second particle

$r_2 = \frac{2m_1\sqrt{2eV/2m_1} }{qB}$

Divide second radius by the first radius;

$\frac{r_2}{r_1} = \frac{2m_1}{m_1} = 2\\\\\frac{R}{r_1} = 2\\\\r_1 = \frac{R}{2}$

Thus, the radius of the arc followed by the lighter is 0.5R.

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