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Question 888: [Modern Physics > Quantum Physics]

Fig.1 gives information on three lines observed in the emission spectrum of hydrogen atoms.

wavelength / nm                    photon energy / 10^–19 J

656                                                      3.03

486                                          .......................................

1880                                                    1.06

Fig.1

(a) Complete Fig.1 by calculating the photon energy for the wavelength of 486 nm.

(b) Fig.2 is a partially completed diagram to show energy levels of a hydrogen atom.

On Fig.2 draw one further labelled energy level, and complete the diagram with arrows to show the energy changes for the other two wavelengths.

Reference: Past Exam Paper – November 2002 Paper 4 Q2

Solution 888:

(a) Energy E = hc / λ = (6.63×10^-19) (3.0×10⁸) / (486×10^-9) = 4.09×10^-19 J

(b)

Energy level drawn at 4.09×10^-19 J

Transition from 4.09×10^-19 to zero should be clear

Transition from 4.09×10^-19 to 3.03×10^-19 should be clear

Question 889: [Current of Electricity]

Deduce the resistance R of the resistor shown in the circuit loop using Kirchhoff’s second law.

Reference: ???

Solution 889:

Note that the circuit is not shown completely. At a junction (represented by a bold dot), the current may change.

A current of 0.5A flows through resistance R. This current of 0.5A also flows through the 10Ω resistor just before resistor R (on the top-right side).

Also, a current of 0.2A flows through the other 10Ω resistor (on the bottom-right). This current of 0.2A also flows through the 20Ω resistor (on the bottom-left) since there is no junction when considering the part 20Ω resistor - battery - 10Ω resistor [the 2 resistors can be considered to be in series with the battery. It is known that the current in a series connection is the same. Current may also change in the connection is in parallel - This is what occurs at the junctions].

The flows of current from the 2 batteries are in different directions (so, we need to subtract the values of the e.m.f.’s from each other) and since the current from the 30V battery is greater than that from the 10V battery, the overall flow of current is anticlockwise.

(Remember again that current changes at the junctions.)

So, from Kirchhoff’s second law, the sum of p.d.’s in a loop is equal to the e.m.f in the loop.

Ohm’s law: p.d V across a resistor = IR

30 – 10 = 0.5R + 0.5(10) + 0.2(10) + 0.2(20)

20 = 0.5R + 11

Resistance R = (20 – 11) / 0.5 = 18 Ω

Question 890: [Options > Telecommunications]

Railway tracks provide a convenient route for communication cables. However, passing electric trains produce interference (noise) of power 7.3 ×10^–5 W in a certain co-axial cable.

Signal-to-noise ratio in this cable must not fall below 25 dB for the effective transmission of the signal.

(a) Show that minimum effective signal power in the cable is 0.023W.

(b) The cable has a loss of 4.8 dB km^–1. Calculate maximum length of cable which can be used without the need for repeater amplifiers for an input signal of power 5.8W.

(c) Co-axial cables are being replaced by optic fibres along railway tracks. Suggest two reasons why this is being done.

Reference: Past Exam Paper – June 2002 Paper 6 Q15

Solution 890:

(a)

{The question states that the signal-to-noise ratio in this cable must not fall below 25 dB.  That is, the ratio of Signal / Noise. This should be the argument of the log function in our formula. Power of noise = 7.3 ×10^–5 W.}

Number of dB = 10 lg(P₁ / P₂)

25 = 10 lg(P / [7.3×10^-5])

Minimum effective signal power P = 0.023 W

(b)

{Input signal power = 5.8W. Minimum effective (output) signal power = 0.023W. The ratio of input signal power / output signal power in the equation would give the total power loss in the cable.}

Change in signal power = 10 lg (5.8 / 0.023) = 24 dB

{The cable has a loss of 4.8 dB km^–1.}

Maximum length of cable = 24 / 5.8 = 5.0 km

(c)

Example:

Less interference

Greater uninterrupted length

No cross-talk

Question 891: [Work, Energy and Power]

When a horizontal force F is applied to a frictionless trolley over distance s, the kinetic energy of the trolley changes from 4 J to 8 J.

If a force of 2F is applied to the trolley over a distance of 2s, what will the original kinetic energy of 4 J become?

A 16 J                          B 20 J                          C 32 J                          D 64 J

Reference: Past Exam Paper – November 2011 Paper 12 Q15

Solution 891:

Answer: B.

Initial kinetic energy of trolley = 4J

Consider the 1^st case.

We need to consider the initial 4J here.

Final kinetic energy = 8J

Work done by force F (= Fs) = 8 – 4 = 4J

So, Fs = 4J

Consider the 2^nd case.

Work done by force 2F = (2F) (2s) = 4(Fs) = 4(4) = 16J since Fs = 4J

Total kinetic energy = 4 + 16 = 20J