Atom 1 of mass 35uand atom 2 of mass 37u are both singly ionized with a charge of +e. After being introduced into a mass spectrometer (Fig. 28-12) and accelerated from rest through a potential difference V=7.3kV, each ion follows a circular path in a uniform magnetic field of magnitude B=0.50T. What is the distance$\mathit{\Delta }\mathit{x}$between the points where the ions strike the detector?

The Mass Of atom 1,m₁=35u.

The Mass of atom 2, m₂=37u.

Both the masses are slightly ionized with a charge of +e.

The potential difference is V=7.3kV.

The uniform magnetic field of magnitude B=0.50T.

By using the equations for the radius of the circular path and the speed of the ion, we can find the equation for$\Delta m$and by using this equation, we can find the distance $\Delta x$between the points where the ions strike the detector.

Formula:

The radius of the circular path, r=mv/qB

The speed,$v=\surd (2qV/m)$

From equation 28-16, the radius of the circular path is

$r=\frac{mv}{qB}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

From equation 28-22, the speed of the ion is

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

Substituting this value in equation (1),

localid="1662959036682" $r=\frac{m}{qB}\sqrt{\frac{2qV}{m}}$

Simplifying further,

localid="1662959087996" $r=\sqrt{\frac{2qV}{q{B}^2}}$

Since,

$x=2r=2\sqrt{\frac{2qV}{q{B}^2}}=\sqrt{\frac{8mV}{q{B}^2}}$

Rearranging this equation for m, we get,

$m=\frac{B^{2}q{x}^2}{8V}$

From this we have

localid="1662959302850" $\Delta m=\left(\frac{B^{2}q}{8V}\right)\left(2x\Delta x\right)$

Substituting the value of x,

localid="1662959354410" width="186">Δm=B2q8V8mVqB2Δx

$\Delta m=B\left(\sqrt{\frac{mq}{2V}}\right)\Delta x$

Thus, the distance between the spots made on the photographic plate is