Lucile C. GCSE Physics tutor, IB Physics tutor, A Level Physics tutor...

Lucile C.

£20 - £22 /hr

Currently unavailable: for new students

Studying: Mathematical Physics (Bachelors) - Edinburgh University

4.9
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17 reviews| 21 completed tutorials

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About me

About me

Hi! I’m a third year student at the University of Edinburgh studying Mathematical Physics. I also completed my IB (International Baccalaureate) with 42 points with 7s in my Higher Level Physics and Maths courses and a 6 in my Higher Level Chemistry. So I can help you with any specific IB problems you may have. However as understanding is the key to success in Maths, Physics and Chemistry I can help you with A-levels and GSCEs too.

Experience

I have been tutoring for 3 years for students with varying levels of ability. My students find me a relaxed, friendly and relatable person. I am no genius, my success has stemmed from good quality and focused studying, and I can show you how to study successfully too. Your exams may look daunting now but hopefully after a few sessions with me you’ll be eager to showcase your hard work.

About me

Hi! I’m a third year student at the University of Edinburgh studying Mathematical Physics. I also completed my IB (International Baccalaureate) with 42 points with 7s in my Higher Level Physics and Maths courses and a 6 in my Higher Level Chemistry. So I can help you with any specific IB problems you may have. However as understanding is the key to success in Maths, Physics and Chemistry I can help you with A-levels and GSCEs too.

Experience

I have been tutoring for 3 years for students with varying levels of ability. My students find me a relaxed, friendly and relatable person. I am no genius, my success has stemmed from good quality and focused studying, and I can show you how to study successfully too. Your exams may look daunting now but hopefully after a few sessions with me you’ll be eager to showcase your hard work.

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About my sessions

I keep the sessions relaxed and stress free, with me no question is stupid. In fact the more questions the better! Each student has their own strengths and weaknesses and we will identify and address yours in our sessions.The material in maths, physics and chemistry courses is progressive and builds upon itself so I first focus on a sound understanding of the basics. Basics don’t have to be boring though! I always manage to find an engaging and creative way of explaining material that students often lack in conventional classrooms. Our sessions will: increase your confidenceencourage you to explain material and guide you through past exam questions.

I keep the sessions relaxed and stress free, with me no question is stupid. In fact the more questions the better! Each student has their own strengths and weaknesses and we will identify and address yours in our sessions.The material in maths, physics and chemistry courses is progressive and builds upon itself so I first focus on a sound understanding of the basics. Basics don’t have to be boring though! I always manage to find an engaging and creative way of explaining material that students often lack in conventional classrooms. Our sessions will: increase your confidenceencourage you to explain material and guide you through past exam questions.

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Ratings & Reviews

4.9from 17 customer reviews
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Natacha (Parent)

September 17 2017

We are very happy with Lucile and would highly recommend her. She is flexible and organised and my daughter who has struggled with pre IB physics is now getting As and Bs. She is friendly and helpful and plans her lessons well. I don't know what would we do without her.

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Isabella (Student)

August 27 2017

very kind and helpful

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Solveig (Student)

November 5 2017

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Solveig (Student)

October 28 2017

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Qualifications

SubjectQualificationGrade
Maths Higher LevelInternational Baccalaureate (IB) (HL)7
Physics Higher Level International Baccalaureate (IB) (HL)7
English Literature Higher Level International Baccalaureate (IB) (HL)7
Chemistry Higher Level International Baccalaureate (IB) (HL)6
German Standard Level International Baccalaureate (IB) (SL)7
Economics Standard LevelInternational Baccalaureate (IB) (SL)6

General Availability

Before 12pm12pm - 5pmAfter 5pm
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fridays
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Subjects offered

SubjectQualificationPrices
ChemistryGCSE£20 /hr
MathsGCSE£20 /hr
PhysicsGCSE£20 /hr
MathsIB£22 /hr
PhysicsIB£22 /hr

Questions Lucile has answered

How can I derive the energy of an electron using the electron in a box model?

The electron in a box model uses the idea that the energy of the electron and that of a standing wave in a box are analogous. So using your knowledge of standing waves you can derive an equation for the electrons energy.

The equation you need to derive is KE = (n2h2)/(8mL2)

It looks a little daunting and is not something I would memorise. Thankfully the derivation just requires 3 main steps and 3 main equations.

Treat the electron in a box as you would treat a standing wave on a string length L. The boundary conditions are that the wave has nodes at either end of the string. If you start drawing out the possible wavelengths on a string length L you will start to see a relationship emerging between the number of anti-nodes, n, and the wavelength, λ. Thus the allowed wavelengths are λ = 2L / n where n = 1, 2, 3, ...

Now that we have this relation we can use our equations from quantum physics, look for one that might includes λ, in this case λ = h / p.

Now substitue p = mv we have λ = 2L / n  = h / (mv). Which rearranges to v = (hn) / (2Lm). So now we have an equation for the velocity of the electron in terms of L.

Remember we needed to find an expression for the energy, but there is the equation KE = mv/ 2. So by substitution:

KE = (n2h2)/(8mL2)

This equation can model the energies of an electron according to it's energy level in the atom. So the n in the equation represents its energy level, the h is Planck's constant, m is the mass of the electron and L is the the length of the box or the space to which the electron is confined.

The electron in a box model uses the idea that the energy of the electron and that of a standing wave in a box are analogous. So using your knowledge of standing waves you can derive an equation for the electrons energy.

The equation you need to derive is KE = (n2h2)/(8mL2)

It looks a little daunting and is not something I would memorise. Thankfully the derivation just requires 3 main steps and 3 main equations.

Treat the electron in a box as you would treat a standing wave on a string length L. The boundary conditions are that the wave has nodes at either end of the string. If you start drawing out the possible wavelengths on a string length L you will start to see a relationship emerging between the number of anti-nodes, n, and the wavelength, λ. Thus the allowed wavelengths are λ = 2L / n where n = 1, 2, 3, ...

Now that we have this relation we can use our equations from quantum physics, look for one that might includes λ, in this case λ = h / p.

Now substitue p = mv we have λ = 2L / n  = h / (mv). Which rearranges to v = (hn) / (2Lm). So now we have an equation for the velocity of the electron in terms of L.

Remember we needed to find an expression for the energy, but there is the equation KE = mv/ 2. So by substitution:

KE = (n2h2)/(8mL2)

This equation can model the energies of an electron according to it's energy level in the atom. So the n in the equation represents its energy level, the h is Planck's constant, m is the mass of the electron and L is the the length of the box or the space to which the electron is confined.

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1 year ago

723 views

How do I derive the indefinite integral of sine?

The integral of sine is pretty easy to remember as it is -cos + C. However you need to be able to prove this, without using the integral of cosine. This method uses sines exponential form.

As eiθ​ = cosθ + isinθ, sine can be expressed as sinθ = (eiθ​- e-iθ​) / 2i. This can make the integration easier as the indefinite integral of ekx = (1/k)ekx and the indefinite integral of e-kx = (-1/k)e-kx

Thus ∫sinx dx = ∫(eix- e-ix) / 2i dx = (1/2i)[ ∫eixdx - ∫e-ix dx] = (1/2i)[eix/i + e-ix​/i] + C =  [-(eix​ + e-ix) / 2]  + C.

Now just as sine can be expressed using complex numbers so can cosine such that cosθ = (eiθ​ + e-iθ​) / 2.

Thus  ∫sinx dx = [-(eix​ + e-ix) / 2]  + C = - cosx + C

The integral of sine is pretty easy to remember as it is -cos + C. However you need to be able to prove this, without using the integral of cosine. This method uses sines exponential form.

As eiθ​ = cosθ + isinθ, sine can be expressed as sinθ = (eiθ​- e-iθ​) / 2i. This can make the integration easier as the indefinite integral of ekx = (1/k)ekx and the indefinite integral of e-kx = (-1/k)e-kx

Thus ∫sinx dx = ∫(eix- e-ix) / 2i dx = (1/2i)[ ∫eixdx - ∫e-ix dx] = (1/2i)[eix/i + e-ix​/i] + C =  [-(eix​ + e-ix) / 2]  + C.

Now just as sine can be expressed using complex numbers so can cosine such that cosθ = (eiθ​ + e-iθ​) / 2.

Thus  ∫sinx dx = [-(eix​ + e-ix) / 2]  + C = - cosx + C

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1 year ago

674 views

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