QCE Physics: Quantum theory Q144

Q144 · Practice questionComplex unfamiliar11 marks

QUESTION 144 (11 marks)

In the Bohr model of the hydrogen atom, a single electron moves at constant speed

v

in a circular orbit of radius

r

around a stationary proton. The electric force between the proton and the electron provides the centripetal force.

In the

n

th allowed orbit, the angular momentum of the electron satisfies

mvr = \frac{nh}{2\pi}

where

m

is the mass of the electron and

n = 1, 2, 3, \dots

Ignore gravitational and relativistic effects. Use data from the formula and data book.

Determine an expression for

v^2

in terms of

k

e

m

and

r

[2 marks]

Determine an expression for the radius

r_n

of the

n

th orbit in terms of

n

h

k

m

and

e

[3 marks]

Calculate the radius of the first orbit (

n = 1

). Show your working.

[2 marks]

The orbital period of the electron is

T_1

for

n = 1

and

T_3

for

n = 3

d(i)

Determine the ratio

\dfrac{T_3}{T_1}

[2 marks]

d(ii)

Kepler's third law states that

\dfrac{T^2}{r^3}

is constant for objects orbiting the same central body. Use your answers to parts b) and d)i) to show that the electron's orbits obey this relationship, and explain why an electron orbiting a proton would be expected to do so.

[2 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

11 marks

ANSWER

(a)

v^2=ke^2/(mr)

. (b)

r_n=n^2h^2/(4\pi^2kme^2)

. (c)

5.30\times10^{-11}\ \mathrm{m}

. (d)(i)

27

; (ii)

T^2/r^3

is constant because the force is inverse-square.

Worked solution

Key insight. Combining the electric force with Bohr's quantisation condition fixes the size of the atom from constants alone. And because the electric force is an inverse-square force, Bohr orbits obey Kepler's third law just as planets do.

Part a · 2 marks

The electric force provides the centripetal force:

\frac{ke^2}{r^2} = \frac{mv^2}{r} \quad\Rightarrow\quad v^2 = \frac{ke^2}{mr}

equates the electric force between the electron and proton with the centripetal force

[1 mark]

rearranges correctly to obtain

v^2 = \dfrac{ke^2}{mr}

[1 mark]

Part b · 3 marks

From

mvr = \dfrac{nh}{2\pi}

v = \dfrac{nh}{2\pi mr}

, so

v^2 = \dfrac{n^2h^2}{4\pi^2m^2r^2}

. Equating with part a):

\frac{n^2h^2}{4\pi^2m^2r^2} = \frac{ke^2}{mr} \quad\Rightarrow\quad r_n = \frac{n^2h^2}{4\pi^2kme^2}

rearranges the quantisation condition to give

v

(or

v^2

) in terms of

r

[1 mark]

equates this with the expression from part a)

[1 mark]

rearranges to obtain the correct expression for

r_n

[1 mark]

Part c · 2 marks

r_1 = \frac{h^2}{4\pi^2 kme^2} = \frac{(6.626\times10^{-34})^2}{4\pi^2(9\times10^{9})(9.109\times10^{-31})(1.6\times10^{-19})^2} = 5.30\times10^{-11}\ \text{m}

Tolerance:

5.2

–

5.4 \times 10^{-11}\ \text{m}

(the accepted Bohr radius is

5.29 \times 10^{-11}\ \text{m}

; the small difference arises from the rounded constants in the data book).

substitutes the correct values of

h

k

m

and

e

[1 mark]

calculates the radius

[1 mark]

Part d(i) · 2 marks

T = \dfrac{2\pi r}{v}

. From part b),

r_n = n^2 r_1

, so

r_3 = 9\,r_1

. From the quantisation condition,

v = \dfrac{nh}{2\pi m r_n} = \dfrac{v_1}{n}

, so

v_3 = \dfrac{v_1}{3}

\frac{T_3}{T_1} = \frac{2\pi r_3 / v_3}{2\pi r_1 / v_1} = \frac{9}{1/3} = 27

Numerical check:

T_1 = 1.52 \times 10^{-16}\ \text{s}

T_3 = 4.11 \times 10^{-15}\ \text{s}

(also

v_1 = 2.18 \times 10^{6}\ \text{m s}^{-1}

, about 0.7% of

c

, so ignoring relativistic effects is reasonable).

recognises

T = \dfrac{2\pi r}{v}

and uses

r_3 = 9r_1

[1 mark]

determines

v_3 = \dfrac{v_1}{3}

and calculates the ratio

[1 mark]

Part d(ii) · 2 marks

\left(\dfrac{T_3}{T_1}\right)^2 = 27^2 = 729

and

\left(\dfrac{r_3}{r_1}\right)^3 = 9^3 = 729

, so

\dfrac{T^2}{r^3}

is the same for

n = 1

and

n = 3

The electric force, like the gravitational force, is an inverse-square force (

F \propto 1/r^2

). When an inverse-square force provides the centripetal force,

\dfrac{mv^2}{r} \propto \dfrac{1}{r^2}

, so

v^2 \propto \dfrac{1}{r}

and

T^2 = \dfrac{4\pi^2r^2}{v^2} \propto r^3

— the same reasoning that gives Kepler's third law for planets.

shows that

(T_3/T_1)^2 = (r_3/r_1)^3\ (= 729)

[1 mark]

explains that both electric and gravitational forces are inverse-square forces providing the centripetal force, leading to

T^2 \propto r^3

[1 mark]

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Quantum theory — Question 144

QUESTION 144 (11 marks)

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