Quantum theory — Question 105

Original QCE Vault practice · 8 marks

Q105 · Practice questionComplex unfamiliar8 marks

QUESTION 105 (8 marks)

An electron enters a uniform magnetic field

B=2.00\times10^{-3}\,\mathrm{T}

directed into the page, with velocity perpendicular to the field. The measured circular-path radius is

2.50\times10^{-2}\,\mathrm{m}

. The speed is low enough that classical momentum

p=mv

may be used.

Electron entering an into-page magnetic field and curving downward in a circular path

Using magnetic force and centripetal force, show that

p=eBr

[2 marks]

Calculate the electron's de Broglie wavelength.

[3 marks]

The accelerating potential difference is doubled while

B

is unchanged. Determine the factor by which the orbital radius and de Broglie wavelength change. Explain.

[3 marks]

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WORKED SOLUTION

Practice marking scheme

8 marks

ANSWER

(a)

p=eBr

; (b)

8.28\times10^{-11}\,\mathrm{m}

; (c)

r'/r=\sqrt2

\lambda'/\lambda=1/\sqrt2

Worked solution

a) Momentum relation

Answer:

p=eBr

The magnetic force provides the centripetal force. Since

v\perp B

e vB=\frac{mv^2}{r}.

Cancelling one factor of

v

mv=eBr.

Since

p=mv

\boxed{p=eBr}.

b) de Broglie wavelength

Answer:

8.28\times10^{-11}\text{ m}

First,

p=eBr =(1.60\times10^{-19})(2.00\times10^{-3})(2.50\times10^{-2}) =8.00\times10^{-24}\text{ kg m s}^{-1}.

Then

\lambda=\frac{h}{p} =\frac{6.626\times10^{-34}}{8.00\times10^{-24}} =8.28\times10^{-11}\text{ m}.

c) Doubling the accelerating potential difference

Answer:

\boxed{r\text{ increases by a factor }\sqrt2}

and

\boxed{\lambda\text{ decreases by a factor }\frac{1}{\sqrt2}}.

An electron accelerated from rest through potential difference

V

gains kinetic energy

eV=E_k=\frac{p^2}{2m}.

Therefore

p\propto\sqrt V

. Doubling

V

makes the momentum

\sqrt2

times larger.

With

B

unchanged,

r=\frac{p}{eB}\propto p,

so the radius increases by

\sqrt2

. Also,

\lambda=\frac{h}{p}\propto\frac1p,

so the wavelength decreases to

1/\sqrt2

of its original value.

Alternative approach: show

v\propto\sqrt V

, then use

r=mv/(eB)

and

\lambda=h/(mv)

Key insight: the same momentum inferred from the magnetic curvature also controls the de Broglie wavelength.

equates magnetic force and centripetal force correctly.

[1 mark]

rearranges to

p=eBr

[1 mark]

determines

p=eBr

[1 mark]

applies

\lambda=h/p

[1 mark]

obtains

8.28\times10^{-11}\text{ m}

, with units.

[1 mark]

establishes

p\propto\sqrt V

from electrical energy and kinetic energy.

[1 mark]

determines

r' / r=\sqrt2

[1 mark]

determines

\lambda'/\lambda=1/\sqrt2

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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Quantum theory — Question 105

QUESTION 105 (8 marks)

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