A uniform magnetic field with a slit system as shown in figure is to be used as momentum filter for

Q: Why is the radius the same for both particles?

Because the positions of the source and the detector are fixed. The particles must travel the exact same semi-circular path to successfully pass through the slit system.

Q: How do we know the mass and charge of the particles?

An alpha particle is a helium nucleus ($^4_2\text{He}$), so it has 2 protons and 2 neutrons (mass $\approx 4\text{ amu}$, charge $= +2e$). A deuteron is a hydrogen-2 nucleus ($^2_1\text{H}$), with 1 proton and 1 neutron (mass $\approx 2\text{ amu}$, charge $= +1e$).

Q: Why do we use kinetic energy instead of velocity?

The problem provides the energy of the alpha particle and asks for the energy of the deuteron. It's mathematically more direct to substitute momentum $p = \sqrt{2mK}$ into the radius formula.

Q: What happens if the magnetic field isn't changed?

If the magnetic field remained $B$, the deuteron would need a different kinetic energy to maintain the same radius. Since $R \propto \frac{\sqrt{mK}}{qB}$, changing $B$ requires a corresponding change in $K$ to keep $R$ constant.

Q: Can this filter separate particles of the same mass but different charges?

Yes! If two particles have the same mass and kinetic energy but different charges, they will have different radii. Only the particle with the specific radius matching the detector's position will pass through.

Question

A uniform magnetic field with a slit system as shown in figure is to be used as momentum filter for high-energy charged particles. With a field $B$ Tesla, it is found that the filter transmits $\alpha$-particles each of energy $5.3	ext{ MeV}$. The magnetic field is increased to $2.3 B$ Tesla and deuterons are passed into the filter. The energy of each deuteron transmitted by the filter is ......MeV.

Sigma Percentile · Accepted Answer

### The Momentum Filter

Imagine a charged particle entering a uniform magnetic field. Instead of moving in a straight line, the magnetic force acts perpendicular to its velocity, bending its path into a perfect circle. A momentum filter uses this exact principle! By placing a source and a detector at fixed positions, only particles that travel in a semi-circle of a very specific radius can successfully pass through the slit system.

In our problem, both the $\alpha$-particle and the deuteron must travel the exact same path to hit the detector. This means their radii must be identical:

$$R_\alpha = R_d$$

### The Master Equation

We know the radius of a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$. However, the problem gives us information about the particles' kinetic energies, not their velocities.

We can bridge this gap by expressing momentum $p = mv$ in terms of kinetic energy $K$. Since $K = \frac{p^2}{2m}$, we have $p = \sqrt{2mK}$. Substituting this into our radius formula gives us the master equation for this problem:

$$R = \frac{\sqrt{2mK}}{qB}$$

### Comparing the Particles

Before we equate the radii, let's establish the relationship between the mass and charge of an $\alpha$-particle and a deuteron.

*   An **$\alpha$-particle** is a helium nucleus ($^4_2	ext{He}$). It contains 2 protons and 2 neutrons. Its mass is approximately $4m_p$ and its charge is $+2e$.
*   A **deuteron** is a hydrogen-2 nucleus ($^2_1	ext{H}$). It contains 1 proton and 1 neutron. Its mass is approximately $2m_p$ and its charge is $+1e$.

From this, we can clearly see that the $\alpha$-particle is twice as massive and has twice the charge of the deuteron:

$$m_\alpha = 2m_d \quad 	ext{and} \quad q_\alpha = 2q_d$$

### The Final Calculation

Now, let's equate the radii expressions for both particles and substitute our known relationships:

$$\frac{\sqrt{2m_\alpha K_\alpha}}{q_\alpha B_\alpha} = \frac{\sqrt{2m_d K_d}}{q_d B_d}$$

Substituting $m_\alpha = 2m_d$, $q_\alpha = 2q_d$, $K_\alpha = 5.3 	ext{ MeV}$, $B_\alpha = B$, and $B_d = 2.3B$:

$$\frac{\sqrt{2(2m_d)(5.3)}}{(2q_d) B} = \frac{\sqrt{2(m_d) K_d}}{q_d (2.3 B)}$$

We can elegantly cancel out the common terms $m_d$, $q_d$, and $B$ from both sides:

$$\frac{\sqrt{2 	imes 5.3}}{2} = \frac{\sqrt{K_d}}{2.3}$$

To isolate $K_d$, we square both sides of the equation:

$$\frac{2 	imes 5.3}{4} = \frac{K_d}{(2.3)^2}$$

$$\frac{5.3}{2} = \frac{K_d}{5.29}$$

Finally, solving for $K_d$:

$$K_d = \frac{5.3 	imes 5.29}{2} = \mathbf{14.0185 	ext{ MeV}}$$

This is the exact kinetic energy the deuteron needs to navigate the increased magnetic field and successfully reach the detector!

Enter Numerical Value:

Visualized Solution

$R_{α} = R_{d} = R$

$R = \frac{2 m K}{q B}$

$\frac{2 m _{α} K _{α}}{q _{α} B _{α}} = \frac{2 m _{d} K _{d}}{q _{d} B _{d}}$

$m_{α} = 2 m_{d}, q_{α} = 2 q_{d}$

$\frac{2 ( 2 m _{d} ) ( 5.3 )}{( 2 q _{d} ) B} = \frac{2 ( m _{d} ) K _{d}}{q _{d} ( 2.3 B )}$

$\frac{2 \times 5.3}{2} = \frac{K _{d}}{2.3}$

$\frac{2 \times 5.3}{4} = \frac{K _{d}}{( 2.3 ) ^{2}}$

$K_{d} = 14.0185 MeV$

$If B_{d} = B, then K_{d} = ?$

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Momentum Filter

The Master Equation

Comparing the Particles

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rα​=Rd​=R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=qB2mK​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } qα​Bα​2mα​Kα​​​=qd​Bd​2md​Kd​​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mα​=2md​,qα​=2qd​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (2qd​)B2(2md​)(5.3)​​=qd​(2.3B)2(md​)Kd​​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 22×5.3​​=2.3Kd​​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 42×5.3​=(2.3)2Kd​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kd​=14.0185 MeV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If Bd​=B, then Kd​=?