# Physics 9702 Doubts | Help Page 181

__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}) (3.0×10

^{8}) / (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(P

_{1 }/ P_{2})
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

This comment has been removed by the author.

ReplyDeleteFor 4/M/J/03 Q.1(c), see solution 892 at

Deletehttp://physics-ref.blogspot.com/2015/07/physics-9702-doubts-help-page-182.html

For 04/O/N/06 Q.6(b),(c)(i),(iii), see solution 895 at

http://physics-ref.blogspot.com/2015/07/physics-9702-doubts-help-page-182.html

Thanks so much!

ReplyDeleteWhy is kinetic energy decreasing in second case of solution 891?& not increasing like the first case?

ReplyDelete???? it is not decreasing??

DeleteSaved my life. Wish my teachers could explain like this.

ReplyDeleteHave you done more recent papers (2016-17)?

Some of them. but they are scattered throughout the site. if you have a specific problem, you may refer me to the question

Delete