Photo AI

Read the following passage and answer the accompanying questions - Leaving Cert Physics - Question 10 - 2007

Question icon

Question 10

Read-the-following-passage-and-answer-the-accompanying-questions-Leaving Cert Physics-Question 10-2007.png

Read the following passage and answer the accompanying questions. Ernest Walton was one of the legendary pioneers who made 1932 the annus mirabilis of experimental ... show full transcript

Worked Solution & Example Answer:Read the following passage and answer the accompanying questions - Leaving Cert Physics - Question 10 - 2007

Step 1

Draw a labelled diagram to show how Cockcroft and Walton accelerated the protons.

96%

114 rated

Answer

The diagram should include:

  • A labeled anode and cathode.
  • A depiction of a proton being accelerated by a high potential difference (PD).
  • Indicate the lithium target where the two helium nuclei are produced.

Show the relationship:

[ PD = \frac{W}{H} ]

Step 2

What is the velocity of a proton when it is accelerated from rest through a potential difference of 700 kV?

99%

104 rated

Answer

Using the work-energy principle:

[ W = q \cdot V ]

Where:

  • ( q = 1.6022 \times 10^{-19} , C ) (charge of proton)
  • ( V = 700 \times 10^{3} , V )

Calculate the kinetic energy:

[ W = (1.6022 \times 10^{-19}) \cdot (700 \times 10^{3}) = 1.12154 \times 10^{-10} , J ]

Using ( KE = \frac{1}{2}mv^2 ):

[ v = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2 \times 1.12154 \times 10^{-10}}{1.676 \times 10^{-27}}} ]

Calculating: [ v \approx 1.16 \times 10^{7} , m/s ]

Step 3

Write a nuclear equation to represent the disintegration of a lithium nucleus when bombarded with a proton.

96%

101 rated

Answer

The nuclear equation is as follows:

[ \text{Li}^{6}{3} + \text{p}^{1}{1} \rightarrow \text{He}^{4}{2} + \text{He}^{4}{2} +\text{Energy} ]

Step 4

Calculate the energy released in this disintegration.

98%

120 rated

Answer

Calculating using the masses:

  • Mass of reactants = ( 1.1646 \times 10^{-26} + 1.6726 \times 10^{-27} = 1.33186 \times 10^{-26} , kg )
  • Mass of products = ( 2.6643 \times 10^{-27} \times 2 = 5.3286 \times 10^{-27} , kg )

Using ( E = mc^2 ):

[ E = (3.00 \times 10^{8} \text{ m/s})^2 \times (\text{mass difference}) ]

[ E = (3.00 \times 10^{8})^2 \times (1.33186 \times 10^{-26} - 5.3286 \times 10^{-27}) ]

Calculating the energy released in the disintegration.

Join the Leaving Cert students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

Other Leaving Cert Physics topics to explore

;