# IB DP Physics E.4 Fission IB Style Question Bank HL Paper 1

### Question

This interaction between a proton and a pion violates two or more conservation laws.
$\mathrm{p}+\pi^{-} \rightarrow \mathrm{K}^{-}+\pi^{+}$
Quark composition of particles:
$\pi^{-}=\mathrm{du}, \pi^{+}=\mathrm{u} \overline{\mathrm{d}}, \mathrm{K}^{-}=\mathrm{su}, \mathrm{p}=\mathrm{uud}$

Which laws are violated by this interaction?

I. Conservation of charge

II. Conservation of strangeness

III. Conservation of baryon number

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

Ans:C

In this interaction, the initial and final states involve different particles, and we need to consider the conservation of various quantum numbers. Let’s evaluate the conservation laws:

I. Conservation of charge:

•  Initial state: p (proton) has a charge of +1, and π⁻ (pion) has a charge of -1. So, the total charge of the initial state is 1 – 1 = 0.
• Final state: K⁻ (kaon) has a charge of -1, and π⁺ (pion) has a charge of +1. So, the total charge of the final state is -1 + 1 = 0.
• The conservation of charge is not violated in this interaction.

II. Conservation of strangeness:

• Strangeness is not conserved in this interaction. The initial state has a strangeness of 0 (since p and π⁻ have no strangeness), while the final state has a strangeness of -1 (K⁻ has strangeness -1, and π⁺ has strangeness 0).

III. Conservation of baryon number:

• Baryon number is not conserved in this interaction. The initial state has a baryon number of 1 (since p is a baryon), while the final state has a baryon number of 0 (since there are no baryons in the final state).

So, the correct answer is: C. II and III only

Conservation of charge (I) is not violated, but both the conservation of strangeness (II) and the conservation of baryon number (III) are violated in this interaction.

### Question

Three statements about the binding energy are provided.
I. The binding energy is the energy required to completely separate the nucleons.
II. The binding energy is equivalent, in units of energy, to the mass defect when a nucleus is formed from its nucleons.

III. The binding energy is the energy released when a nucleus is formed from its nucleons.
Which statements are true?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

Ans:D

The correct statements are:

I. The binding energy is the energy required to completely separate the nucleons.
II. The binding energy is equivalent, in units of energy, to the mass defect when a nucleus is formed from its nucleons.
III. The binding energy is the energy released when a nucleus is formed from its nucleons.

So the answer is D, all three statements are true.

### Question

The uncertainty in the energy required for pair production is $$3 \mathrm{MeV}$$. What is the uncertainty in the lifetime of the pair?

A. $$10^{-7} \mathrm{~s}$$

B. $$10^{-12} \mathrm{~s}$$

C. $$10^{-17} \mathrm{~s}$$

D. $$10^{-22} \mathrm{~s}$$

Ans:D

The uncertainty principle, as formulated by Werner Heisenberg, states that there is a fundamental limit to how precisely we can simultaneously know certain pairs of physical properties. The most well-known form of the uncertainty principle relates the uncertainty in position ($$\Delta x$$) and the uncertainty in momentum ($$\Delta p$$) of a particle. However, a similar principle applies to the uncertainty in energy and time.

The uncertainty principle for energy and time is given by:

$$\Delta E \cdot \Delta t \geq \frac{\hbar}{2}$$

In your case, you have the uncertainty in energy ($$\Delta E$$) as $$3\, \text{MeV}$$. To find the uncertainty in time ($$\Delta t$$), you can rearrange the equation:

$$\Delta t \geq \frac{\hbar}{2\Delta E}$$

$$\Delta t \geq \frac{1.0545718 \times 10^{-34}\, \text{J·s}}{2 \cdot 3 \times 10^6\, \text{eV}}$$

First, convert the energy from electronvolts (eV) to joules (J):

$$\Delta t \geq \frac{1.0545718 \times 10^{-34}\, \text{J·s}}{2 \cdot 3 \times 10^6 \times 1.602176634 \times 10^{-19}\, \text{J/eV}}$$

$$\Delta t \geq \frac{1.0545718 \times 10^{-34}\, \text{J·s}}{6.0122598 \times 10^{-13}\, \text{J}}$$

Now, calculate the value:

$$\Delta t \geq 1.75356 \times 10^{-22}\, \text{s}$$

So, the uncertainty in the lifetime of the pair is approximately $$1.75356 \times 10^{-22}$$ seconds, which is closest to option D

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