This question is about the isotopes of chromium - AQA - A-Level Chemistry - Question 2 - 2020 - Paper 1

Let the abundance of 54Cr be x% and the abundance of 53Cr be y%. Given that:

• Total abundance = 100%
• Abundance of 52Cr = 86.1%

We can set up the equation:

$x + 86.1 + y = 100$

Simplifying, we find: $x + y = 13.9$

Next, using the weighted average formula for atomic mass:

$52.1 = (x)(54) + (86.1)(52) + (y)(53)$

Solving these two equations gives:

• Abundance of 54Cr = 3.9%
• Abundance of 53Cr = 10.0%

Similarity: Both 54Cr and 52Cr have the same number of protons which define them as chromium atoms.

Difference: 54Cr has 2 more neutrons compared to 52Cr, resulting in a different nucleon number.

1. Ionization is necessary to allow the isotopes to be accelerated by an electric field.
2. Ionization allows the isotopes to be influenced by magnetic fields, which are essential for their separation based on mass-to-charge ratios.

To find time (t), we first derive v from the kinetic energy (KE) equation:

$KE = \frac{mv^2}{2}$ We rearrange to find: $v = \sqrt{\frac{2 \cdot KE}{m}}$ Given:

• KE = 1.102 × 10^-1 J
• Mass of 52Cr ion, m = 52 / (6.022 × 10²³) kg Using this mass, calculate v:

1. Calculate m: $m = \frac{52}{6.022 \times 10^{23}} \approx 8.63 \times 10^{-26} \text{ kg}$
2. Calculate v using the KE: $v = \sqrt{\frac{2 \cdot (1.102 \times 10^{-1})}{8.63 \times 10^{-26}}} \approx 1.23 \times 10^{13} \text{ m/s}$
3. Now apply the formula for time using distance (d = 1.25 m): $t = \frac{d}{v} = \frac{1.25}{1.23 \times 10^{13}} \approx 1.02 \times 10^{-14} \text{ s}$