A different nucleus can be formed by bombarding a stable nucleus with an energetic α-particle.

Question 14

A different nucleus can be formed by bombarding a stable nucleus with an energetic α-particle.

²³₁₁Na is bombarded with an energetic α-particle.

What could be the products of this nuclear reaction?

A ²⁵₁₀Ne + neutron

B ²⁵₁₁Na + proton

C ²⁶₁₂Mg + β

D ²⁷₁₃Al + γ

Reference: Past Exam Paper – November 2012 Paper 12 Q40

Solution:

Answer: D.

An α-particle is a ⁴₂He nucleus.

For any nuclear reaction, the proton number and the nucleon number should be conserved. That is, the proton number and nucleon number of the reactants should be equal to that of the products.

²³₁₁Na   +   ⁴₂He           - - - >

Proton number of reactants = 11 + 2 = 13

Nucleon number of reactants = 23 + 4 = 27

Consider choice A: (A neutron is ¹₀n)

Proton number = 10 + 0 = 10 [not equal]

Nucleon number = 25 + 1 = 26 [not equal]

Consider choice B: (A proton is ¹₁p)

Proton number = 11 + 1 = 11 [not equal]

Nucleon number = 25 + 1 = 26 [not equal]

Consider choice C: (A beta particle is ⁰_-1β)

Proton number = 12 + -1 = 11 [not equal]

Nucleon number = 26 + 0 = 26 [not equal]

Consider choice D: (A gamma ray is ⁰₀γ)

Proton number = 13 + 0 = 13 [equal]

Nucleon number = 27 + 0 = 27 [equal]

In the circuit shown, the reading on the ammeter is zero. The four resistors have different resistances R1, R2, R3 and R4.

Question 26

In the circuit shown, the reading on the ammeter is zero.

The four resistors have different resistances R1, R2, R3 and R4.

Which equation is correct?

A R1 – R3 = R2 – R4

B R1 × R3 = R2 × R4

C R1 – R4 = R2 – R3

D R1 × R4 = R2 × R3

Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q36

Solution:

Answer: D.

For a current to flow in the ammeter, there should be a potential difference between the terminals of the ammeter. That is, the potential at one terminal should be different than the potential at the other terminal.

Since the ammeter reading is zero, there must be no potential difference across it. That is, the potential at each terminal of the ammeter is the same – there is no difference.

The potential at the negative terminal of the cell may be taken to be zero and the potential at the positive terminal may be taken to be equal to the e.m.f. of the cell.

One approach is to consider that the potential difference across the resistor with resistance R₂ must equal the potential difference across the resistor with resistance R₄.

The circuit consists of 2 branches:

Branch 1: R₁ and R₂

Branch 2: R₃ and R₄

Both branches form a potential divider. So, the p.d. across the resistors can easily be obtained by the potential divider formula.

Let the e.m.f. of the cell be E.

p.d. across R₂ = R₂ × E / (R₁ + R₂)

p.d. across R₄ = R₄ × E / (R₃ + R₄)

Since no current flows, these 2 potentials should be equal.

R₂ × E / (R₁ + R₂) = R₄ × E / (R₃ + R₄)

R₂ / (R₁ + R₂) = R₄ / (R₃ + R₄)

Rearranging this gives R₁R₄ = R₂R₃.