Beta particles may be emitted during the decay of an unstable nucleus of an atom.

Question 19

(a) Give one example of

a hadron:

a lepton:

[1]

(b) Describe, in terms of the simple quark model,

(i) a proton, [1]

(ii) a neutron. [1]

(c) Beta particles may be emitted during the decay of an unstable nucleus of an atom. The emission of a beta particle is due to the decay of a neutron.

(i) Complete the following word equation for the particles produced in this reaction.

neutron → .................................... + .................................... + .................................... [1]

(ii) State the change in quark composition of the particles during this reaction. [1]

 [Total: 5]

Reference: Past Exam Paper – June 2016 Paper 21 Q7

Solution:

(a)

hadron: neutron/proton

and

lepton: electron/(electron) neutrino

(b)

(i) proton: up up down                         or uud

(ii) neutron: up down down     or udd

(c)

(i) neutron → proton + electron + (electron) antineutrino

{Note that the decay of a neutron to produce proton is a known one and can be directly inserted to obtain the answer.

Beta particles are emitted. So, one of the blanks should be beta. Beta particles are the same as electrons. So, we can write electrons instead of beta.

Beta particles / electrons are examples of leptons. In an equation if a lepton is emitted, its associated antimatter should also be emitted so that the overall lepton number is zero.

For leptons, a matter (e.g. electron, neutrino) has a lepton number of +1 while an antimatter (e.g. anti-electron, antineutrino) has a lepton number of -1.

So along with the electron, an antimatter should be emitted – either anti-electron or antineutrino.

If an antielectron is emitted, the overall charge on the left-hand side would be 0 (charge of neutron) while that on the right-hand side would be (+1 – 1 + 1 =) +1 (due to proton, electron and antielectron respectively). The charge is not conserved. It cannot be an antielectron.

So, the other particle is an anti-neutrino.}

(ii)

up down down (quarks) change to up up down (quarks)

or

down (quark) changes to up (quark)

A nuclear reaction occurs when a uranium-235 nucleus absorbs a neutron. The reaction may be represented by the equation

Question 18

(a) A nuclear reaction occurs when a uranium-235 nucleus absorbs a neutron. The reaction may be represented by the equation:

²³⁵₉₂U   +          ^W_X n     - - - >  ⁹³₃₇Rb +         ¹⁴¹_ZCs +          Y^W_X n

State the number represented by the letter

W .............................................................

X ..............................................................

Y ..............................................................

Z ...............................................................

[3]

(b) The sum of the masses on the left-hand side of the equation in (a) is not the same as

the sum of the masses on the right-hand side.

Explain why mass seems not to be conserved. [2]

Reference: Past Exam Paper – June 2012 Paper 22 Q7

Solution:

(a)

W = 1 and X = 0                    

{^W_X n is a neutron. So, W = 1 and X = 0}

Y = 2                                      

Z = 55                                    

{On both sides, the total proton number and nucleon number should be the same.

Proton numbers:

92 + 0 = 37 + Z + (y×0)

Giving Z = 55

Nucleon numbers:

235 + 1 = 93 + 141 + (y×1)

236 = 234 + y

y = 236 – 234 = 2}

(b)

Mass–energy is conserved.

{Some of the mass is converted to energy, such that energy E = Δmc² where Δm is the difference in mass. So, the mass is actually conserved, but it is now on the form of energy.}

Energy is released as gamma or photons or kinetic energy of products or EM radiation.

Some reactions of chromium ions are shown below.

Question 5

(a) Some reactions of chromium ions are shown below.

Reference: Past Exam Paper – 9701 March 2016 Paper 42 Q5

Solution:

(a)

i) any metal with an E^o value more negative than –0.41V, e.g. Fe, Mn, Zn, Mg, Cr, Al

[Observing the net charge on the complex ion, we can calculate the charge on the Cr ion in both complexes. We see that charge reduces from +3 to +2 and hence we need a metal that will itself go through oxidation and act as a reducing agent. Therefore we need a metal whose E^o value is more negative than of Cr⁺³ to Cr⁺²]

ii)

M1: value of E_cell correctly calculated (with correct sign) for metal named in (i)

M2: E^o _cell is positive and so reaction is feasible

[E_cell is calculated by subtracting lower value (oxidation) from a greater value(reduction)]

(b) M1: (Cr₂O₇ ^2– + 14H⁺ + 6e^– ⇌ 2Cr³⁺ + 7H₂O) E^o = +1.33V

 (H₂O₂ + 2H⁺ + 2e^– ⇌ 2H₂O) E^o = +1.77V E^o _cell = 0.44 V

M2: E^o cell (0.44V) is positive (so the reaction is feasible)/E^o (Cr₂O₇ ^2– /Cr³⁺) is less positive than E^o (H₂O₂ /H₂O)

(c)  M1: Cr₂O₇ ^2–: ox.no Cr = +6 because –2 = 2 × ox.no(Cr) + (7 × –2) CrO₄ ^2–: ox.no Cr = +6 because –2 = ox.no(Cr) + (4 × –2) M2: no change in oxidation number, so reaction is not redox

[calculate oxidation numbers of Cr in both ionic states and calculate change in oxidation number(final – initial);

positive change=> oxidation

negative change => reduction

no change=> not a redox reaction]

(d) M1: no. moles Cr deposited = 0.0312/ 52 = 6.0 × 10^–4 moles

M2: deduction that 6 moles of e^– needed per mole of Cr/ reaction is Cr₂O₇ ^2- + 14H⁺ + 12e^– → 2Cr + 7H₂O

[since there is no electrode reaction given for Cr₂O₇ ^2- to Cr(metal), we need to combine two separate reactions to achieve: Cr₂O₇ ^2- + 14H⁺ + 12e^– → 2Cr + 7H₂O. 2 moles of chromium metal requires 12 moles of electrons as seen from above equation, therefore one mole of metal requires 6 moles of electrons. ]

M3: no. moles of e^– = 6 × 6.0 × 10^–4 = (0.125 × t)/ 96 500 so t = (6 × 6.0 × 10^–4 × 96 500)/(0.125 × 60) = 46.3min/ 0.772 h/ 2780s  

Solutions provided by Kashish Varshney, India