A narrow beam of singly charged potassium ions with a kinetic...

The radius of the circular path of an ion in the magnetic field is given by $R = \dfrac{mv}{qB} = \dfrac{\sqrt{2mK}}{qB}$.

Since the width of the magnetic field $d$ is much smaller than $R$, the deflection angle $\theta$ is small. The angle of emergence is $\theta \approx \sin\theta = \dfrac{d}{R}$.

The transverse deflection of the ion inside the magnetic field is $y_1 = R(1 - \cos\theta) \approx R \dfrac{\theta^2}{2} = \dfrac{d^2}{2R}$.

After exiting the magnetic field, the ion travels in a straight line for a distance $L$ to the screen. The additional deflection is $y_2 = L \tan\theta \approx L\theta = \dfrac{Ld}{R}$.

The total deflection on the screen is $y = y_1 + y_2 = \dfrac{d^2}{2R} + \dfrac{Ld}{R} = \dfrac{d}{R}\left(L + \dfrac{d}{2}\right)$.

Let $D = L + \dfrac{d}{2} = 95.5 \text{ cm} + 0.5 \text{ cm} = 96.0 \text{ cm} = 0.96 \text{ m}$.

Substituting $R$, we get $y = \dfrac{d D q B}{\sqrt{2mK}}$.

The mass of the ion is $m = A m_0$, where $m_0 = 1.6 \times 10^{-27} \text{ kg}$ and $A$ is the mass number.

The separation between the two isotopes on the screen is $\Delta y = y_{39} - y_{41} = \dfrac{d D q B}{\sqrt{2 m_0 K}} \left( \dfrac{1}{\sqrt{39}} - \dfrac{1}{\sqrt{41}} \right)$.

Let's calculate the constant factor:
$d = 0.01 \text{ m}$
$D = 0.96 \text{ m}$
$q = 1.6 \times 10^{-19} \text{ C}$
$B = 0.500 \text{ T}$
$K = 32 \text{ keV} = 32 \times 10^3 \times 1.6 \times 10^{-19} \text{ J}$
$2 m_0 K = 2 \times (1.6 \times 10^{-27}) \times (32 \times 10^3 \times 1.6 \times 10^{-19}) = 16.384 \times 10^{-42} \text{ kg}^2 \text{ m}^2 / \text{ s}^2$
$\sqrt{2 m_0 K} = \sqrt{16.384 \times 10^{-42}} \approx 4.048 \times 10^{-21} \text{ kg m/s}$

Now, substituting the values:
$\dfrac{d D q B}{\sqrt{2 m_0 K}} = \dfrac{0.01 \times 0.96 \times 1.6 \times 10^{-19} \times 0.5}{4.048 \times 10^{-21}} = \dfrac{0.768 \times 10^{-21}}{4.048 \times 10^{-21}} \approx 0.1897 \text{ m}$

The difference in the mass factor is:
$\dfrac{1}{\sqrt{39}} - \dfrac{1}{\sqrt{41}} \approx 0.16013 - 0.15617 = 0.00396$

Thus, the separation is:
$\Delta y \approx 0.1897 \times 0.00396 \approx 0.00075 \text{ m} = 0.75 \text{ mm}$.

Answer: $0.75 \text{ mm}$