# IB DP Physics E.1 Structure of the atom IB Style Question Bank HL Paper 1

### Question

Which statement about atomic nuclei is correct? The density is…

A. directly proportional to mass number.

B. inversely proportional to nuclear radius.

C. inversely proportional to volume.

D. constant for all nuclei.

Ans:D

In atomic nuclei, the density is relatively constant across different nuclei. This is because the density of a nucleus is determined by the ratio of its mass to its volume. While the mass and volume of nuclei can vary, the ratio remains relatively constant for most nuclei. So, the correct answer is D

### Question

In the Bohr model for hydrogen, the radius of the electron orbit in the $$\mathrm{n}=2$$ state is four times that of the radius in the $$n=1$$ state.
What is $$\frac{\text { speed of the electron in the } n=2 \text { state }}{\text { speed of the electron in the } n=1 \text { state }}$$ ?

A. $$\frac{1}{4}$$

B. $$\frac{1}{2}$$

C. 2

D. 4

Ans:B

In the Bohr model for hydrogen, the radius $$(r)$$ of the electron orbit is inversely proportional to the principal quantum number $$(n)$$ according to the following relationship:
$r \propto \frac{1}{n^2}$

So, if the radius of the electron orbit in the $$n=2$$ state is four times that of the $$n=1$$ state, we can express it as:
$\frac{r_{n=2}}{r_{n=1}}=4$

In the Bohr model of the hydrogen atom, the speed of the electron $$(v)$$ is indeed inversely proportional to the square root of the radius $$(r)$$ of the electron’s orbit. The correct relationship is:

$v \propto \frac{1}{\sqrt{r}}$

So, if the radius of the electron orbit in the $$n=2$$ state is four times that of the $$n=1$$ state $$r_{n=2}=4 r_{n=1}$$ ), then the speed of the electron in the $$n=2$$ state compared to the $$n=1$$ state is:
$\frac{v_{n=2}}{v}=\frac{1}{\sqrt{4}}=\frac{1}{2}$

### Question

I. Most of the atom is empty space.
II. The positive charge of the atom is concentrated in a small volume.
III. The electrons have discrete energy levels.

Which of these claims can be deduced from the Rutherford-Geiger-Marsden scattering experiment?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

Ans:A

The Rutherford-Geiger-Marsden scattering experiment, commonly known as the Rutherford gold foil experiment, provided important insights into the structure of the atom. Let’s examine each of the claims in light of this experiment:

I. Most of the atom is empty space:

• This claim can be deduced from the Rutherford-Geiger-Marsden experiment. Rutherford’s experiment showed that alpha particles, which are positively charged, mostly passed through the gold foil with little to no deflection, indicating that the atom is mostly empty space. However, a small fraction of the alpha particles did experience significant deflection, suggesting that there is a concentrated positive charge in a small volume within the atom.

II. The positive charge of the atom is concentrated in a small volume:

• This claim can also be deduced from the Rutherford-Geiger-Marsden experiment. The significant deflection of some alpha particles indicated that the positive charge in the atom is concentrated in a small, dense nucleus at the center of the atom.

III. The electrons have discrete energy levels:

• The Rutherford-Geiger-Marsden experiment did not directly provide information about the discrete energy levels of electrons. This aspect of atomic structure is better explained by later developments in atomic theory, such as Bohr’s model of the atom and quantum mechanics.

Based on the information provided by the Rutherford-Geiger-Marsden experiment, claims I and II can be deduced, but claim III cannot. Therefore, the correct answer is:A. I and II only

### Question

The following reaction is proposed for the collision of a proton $$\mathrm{p}$$ and a neutron $$\mathrm{n}$$.
$\mathrm{p}+\mathrm{n} \rightarrow \mathrm{p}+\pi^0$
The neutral pion $$\pi^0$$ consists of an up quark and an anti-up quark.
Which conservation law does this equation violate?

A. Baryon number

B. Charge

C. Lepton number

D. Strangeness

Ans:A

In the reaction, the initial state has a baryon number of 2 (one proton and one neutron), and the final state also has a baryon number of 1 one proton . Baryon number is conserved in this reaction.

### Question

Consider the Feynman diagram below. What is the exchange particle X?

A Lepton

B Gluon

C Meson

D Photon 