1 (Total for Multiple Choice Questions = 10 marks) · 2015-09-23 · P v 3 1 2014 2014 81 2 P aper 2 Mark scheme Question Number Acceptable answers Additional guidance Mark 1 C 1

Pearson Edexcel Level 3 Advanced GCE in Physics Sample Assessment Materials – Issue 1 – October 2014 © Pearson Education Limited 2014

81

Pape

r 2

Mar

k sc

hem

e Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

1 C

1 2

C

1

3 D

1 4

A

1

5 C

1 6

B

1

7 A

1 8

C

1

9 C

1 10

D

1 (T

otal

for

Mul

tiple

Cho

ice

Que

stio

ns =

10

mar

ks)


82

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

11 (a

) •

The

ener

gy e

quiv

alen

t to

the

mas

s def

icit

(1)

• W

hen

nucl

eons

bin

d to

geth

er to

form

an

atom

ic n

ucle

us (1

)

2

11 (b

) •

Cal

cula

tion

of m

ass d

iffer

ence

in k

g (1

) •

Use

of E

= c

2 Δm

(1)

• E

= 2.

77 ×

10-1

1 J (1

)

Exam

ple

of c

alcu

latio

n:

(235

.043

9 +

1.00

8665

) u –

(140

.914

4 +

91.9

262+

(3×1

.008

665)

) u =

0.1

86u

(0.1

860

u ×

1.66

× 1

0-27 k

g) ×

(3 ×

108 m

s-1)2 =

2.

77 ×

10-1

1 J

3

(Tot

al fo

r Q

uest

ion

11 =

5 m

arks

)


83

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

12 (a

) •

Use

of L

= λ

/2 (1

) •

Use

of v

= fλ

(1)

• f =

180

Hz

(1)

Exam

ple

of c

alcu

latio

n:

λ =

2 ×

0.45

m =

0.9

0 m

f =

v/λ

= 1

60 m

s-1/0

.9 m

= 1

78 H

z

3 12

(b)

An

expl

anat

ion

that

mak

es re

fere

nce

to:

Eith

er

• Th

e os

cilla

ting

fram

e ca

uses

the

lead

sphe

res t

o de

form

pl

astic

ally

(1)

• A

nd th

is re

mov

es e

nerg

y fr

om th

e os

cilla

ting

fram

e (1

) •

So th

e am

plitu

de o

f osc

illat

ions

dec

reas

e w

ith ti

me

as

show

n by

the

grap

h (1

)

OR

•

Sphe

res c

ollid

e/vi

brat

e (1

) •

Hen

ce e

nerg

y di

ssip

ated

(1)

• So

the

ampl

itude

of o

scill

atio

ns d

ecre

ase

with

tim

e as

sh

own

by th

e gr

aph

(1)

3

(Tot

al fo

r Q

uest

ion

12 =

6 m

arks

)


84

Qu

esti

on

n

um

ber

A

ccep

tab

le a

nsw

ers

A

dd

itio

nal

gu

idan

ce

Mar

k

13 *

Th

is q

uest

ion

asse

sses

a st

uden

t’s a

bilit

y to

show

a c

oher

ent

and

logi

cally

stru

ctur

ed a

nsw

er w

ith li

nkag

es a

nd fu

lly-

sust

aine

d re

ason

ing.

Mar

ks a

re a

war

ded

for i

ndic

ativ

e co

nten

t and

for h

ow th

e an

swer

is st

ruct

ured

and

show

s lin

es o

f rea

soni

ng.

The

follo

win

g ta

ble

show

s how

the

mar

ks sh

ould

be

awar

ded

for i

ndic

ativ

e co

nten

t. N

umbe

r of

indi

cativ

e m

arki

ng

poin

ts se

en in

an

swer

Num

ber o

f m

arks

aw

arde

d fo

r ind

icat

ive

mar

king

poi

nts

6 4

5–4

3 3–

2 2

1 1

0 0

Gui

danc

e on

how

the

mar

k sc

hem

e sh

ould

be

appl

ied:

Th

e m

ark

for i

ndic

ativ

e co

nten

t sho

uld

be a

dded

to

the

mar

k fo

r lin

es o

f rea

soni

ng. F

or e

xam

ple,

an

answ

er w

ith fi

ve in

dica

tive

mar

king

poi

nts w

hich

is

parti

ally

stru

ctur

ed w

ith so

me

linka

ges a

nd li

nes o

f re

ason

ing

scor

es 4

mar

ks (3

mar

ks fo

r ind

icat

ive

cont

ent a

nd 1

mar

k fo

r par

tial s

truct

ure

and

som

e lin

kage

s and

line

s of r

easo

ning

).

If th

ere

are

no li

nkag

es b

etw

een

poin

ts, t

he sa

me

five

indi

cativ

e m

arki

ng p

oint

s wou

ld y

ield

an

over

all

scor

e of

3 m

arks

(3 m

arks

for i

ndic

ativ

e co

nten

t and

no

mar

ks fo

r lin

kage

s).


85

Qu

esti

on

n

um

ber

A

dd

itio

nal

gu

idan

ce

Mar

k

13*

(con

tinue

d)

The

follo

win

g ta

ble

show

s how

the

mar

ks sh

ould

be

awar

ded

for s

truct

ure

and

lines

of r

easo

ning

.

N

umbe

r of m

arks

aw

arde

d fo

r stru

ctur

e of

ans

wer

and

su

stai

ned

line

of re

ason

ing

A

nsw

er sh

ows a

co

here

nt a

nd lo

gica

l st

ruct

ure

with

lin

kage

s and

fully

su

stai

ned

lines

of

reas

onin

g de

mon

stra

ted

thro

ugho

ut

2

Ans

wer

is p

artia

lly

stru

ctur

ed w

ith so

me

linka

ges a

nd li

nes o

f re

ason

ing

1

Ans

wer

has

no

linka

ges b

etw

een

poin

ts a

nd is

un

stru

ctur

ed

0


86

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

13*

(con

tinue

d)

Indi

cativ

e co

nten

t

• So

und

wav

es in

cide

nt u

pon

surf

aces

with

in th

e co

ncer

t hal

l w

ill b

e re

flect

ed.

•

Som

e fr

eque

ncie

s will

arr

ive

from

diff

eren

t dire

ctio

ns w

ith a

ph

ase

diff

eren

ce o

f (an

y od

d m

ultip

le o

f) π

radi

ans (

1) O

R

path

diff

eren

ce is

odd

num

ber o

f hal

f wav

elen

gths

• D

estru

ctiv

e su

perp

ositi

on/in

terf

eren

ce w

ill o

ccur

, cau

sing

th

e w

aves

with

thos

e fr

eque

ncie

s to

be q

uiet

er th

an o

ther

s.

• O

ther

freq

uenc

ies a

rriv

e w

ith a

pha

se d

iffer

ence

of z

ero

or

(any

mul

tiple

of)

2π

radi

ans (

1) O

R a

who

le n

umbe

r of

wav

elen

gths

• C

onst

ruct

ive

supe

rpos

ition

/inte

rfer

ence

will

occ

ur, c

ausi

ng

wav

es w

ith th

ose

freq

uenc

ies t

o be

loud

er th

an o

ther

s.

•

Prob

lem

aris

es d

ue to

refle

ctio

ns fr

om w

alls

, so

use

abso

rbin

g m

ater

ial o

n su

rfac

es to

redu

ce re

flect

ions

.

●

6


87

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

13*

(con

tinue

d)

Alte

rnat

ive

appr

oach

bas

ed o

n st

andi

ng w

aves

: •

Soun

d w

aves

inci

dent

upo

n su

rfac

es w

ithin

the

conc

ert h

all

will

be

refle

cted

.

• R

efle

ctio

ns fr

om w

alls

set u

p st

andi

ng w

aves

(in

room

)

• N

odes

and

ant

inod

es a

re fo

rmed

for c

erta

in fr

eque

ncie

s of

soun

d

• N

odes

are

are

as o

f zer

o/lo

w a

mpl

itude

so th

e fr

eque

ncie

s of

thos

e so

und

wav

es w

ill b

e qu

iete

r tha

n ot

hers

• A

ntin

odes

are

are

as o

f max

imum

am

plitu

de so

the

freq

uenc

ies o

f the

se so

und

wav

es w

ill b

e lo

uder

than

oth

ers

•

Prob

lem

aris

es d

ue to

refle

ctio

ns fr

om w

alls

, so

use

abso

rbin

g m

ater

ial o

n su

rfac

es to

redu

ce re

flect

ions

(Tot

al fo

r Q

uest

ion

13 =

6 m

arks

)


88

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

14 (a

) •

Use

of

22

11

sinsin

θθ

nn

= (1

) •

c =

67° (1

) •

Det

erm

ines

the

angl

e of

inci

denc

e is

70° (1

) •

so i

> c

so th

e ra

y do

es to

tally

inte

rnal

ly re

flect

(1)

Exam

ple

of c

alcu

latio

n: o

sin90

1.44

sin1.56

×=

×c

c

= 67

.4o

4

14 (b

)(i)

Two

cons

truct

ion

rays

from

: •

ray

from

tip

of o

bjec

t par

alle

l to

prin

cipa

l axi

s dra

wn

then

re

frac

ted

thro

ugh

the

foca

l poi

nt (1

) •

ray

draw

n fr

om ti

p of

obj

ect t

hrou

gh c

entre

of l

ens

(1)

• ra

y dr

awn

from

foca

l poi

nt th

roug

h tip

of o

bjec

t and

then

re

frac

ted

para

llel t

o th

e pr

inci

pal a

xis (

1)

A

nd

• ra

ys e

xten

ded

back

to lo

cate

tip

of im

age

on th

e sa

me

side

as t

he

obje

ct (1

)

Exam

ple

of d

iagr

am:

3


89

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

14 (b

)(ii)

•

Use

of m

= v

/u (1

)

• U

se o

f f

vu

11

1=

+ a

nd su

bstit

utin

g fo

r v o

r u (1

)

• u

= 6.

1 cm

(1)

Exam

ple

of c

alcu

latio

n:

uv

uv5.3

5.3−

=∴

−=

5.815.31

1=

−+

uu

5.81

5.31

5.3=

−∴

u

5.8

5.25.3=

∴u

cm07.6

5.35.2

5.8=

×=

u

3 (T

otal

for

Que

stio

n 14

= 1

0 m

arks

)


90

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

15

• U

se o

f NkT

pV=

(1)

• C

onve

rsio

n of

tem

pera

ture

to K

elvi

n (1

) •

p =

95.4

kPa

(1)

• C

alcu

latio

n of

exc

ess p

ress

ure

(1)

• U

se o

f p =

F/A

(1)

• Δ

F =

995

N (1

) •

Sens

ible

com

men

t, e.

g. th

is is

a la

rge

forc

e so

cou

ld m

ake

the

door

har

d to

ope

n (1

)

Exam

ple

of c

alcu

latio

n:

22

11

TpTp

=

()

() K

22.5

273

K3.3

273

Pa10

102

3

121

2++

××

=×

=TT

pp

Pa10

95.37

3×

=

Δp

= (1

02 −

95.3

7) k

Pa =

6.6

3 kP

a N

994.5

Pa10

6.63

m0.15

32

=×

×=

Δ=

Δp

AF

7 (T

otal

for

Que

stio

n 15

= 7

mar

ks)


91

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

16

• U

se o

f F =

k Δ

x (1

) •

k =

14.4

N m

-1 (1

) •

Use

of

kmT

π2=

(1)

• U

se o

f f =

1/T

(1)

•

f = 2

.4 H

z (1

)

Exam

ple

of c

alcu

latio

n:

k =

mg/Δ

x =

66×

10-3

kg ×

9.81

m s-2

)/4.5

× 1

0-2 m

=

14.4

N m

-1

T =

2π(0

.066

/14.

4.0)

1/2 =

0.4

25 s

f=1/

T =

1/0.

425

= 2.

35 H

z

5

(Tot

al fo

r Q

uest

ion

16 =

5 m

arks

)


92

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

17 (a

) A

n ex

plan

atio

n th

at m

akes

refe

renc

e to

:

• th

e te

mpe

ratu

re is

con

stan

t whe

n th

e pu

ree

boils

bec

ause

the

aver

age

kine

tic e

nerg

y of

the

mol

ecul

es in

the

pure

e is

con

stan

t. (1

) •

whe

n bo

iling

occ

urs,

the

ther

mal

ene

rgy

supp

lied

incr

ease

s the

po

tent

ial e

nerg

y of

the

mol

ecul

es c

ausi

ng th

e m

olec

ules

to m

ove

furth

er a

part

(pro

duci

ng st

eam

) (1)

OR

whe

n bo

iling

occ

urs,

the

ther

mal

ene

rgy

supp

lied

incr

ease

s the

pot

entia

l ene

rgy

of th

e m

olec

ules

bre

akin

g m

olec

ular

bon

ds. (

1)

2

17 (b

) •

Use

of

θΔ

=Δ

mc

Ew

ith a

tem

pera

ture

cha

nge

of 8

0° C (1

)

• 1

13

CkgJ

103.94

−−

×=c

(1)

Exam

ple

of c

alcu

latio

n:

Tem

pera

ture

rise

= (1

01 –

21)

° C

() C

21101

kg0.444

J175000

−×

×=

c

11

3C

kgJ10

3.94

−−

×=c

2

17 (c

) •

Use

of

Lm

E=

Δ (1

) •

Con

vert

peak

vol

tage

and

cur

rent

to r.

m.s.

val

ues (

230

V a

nd

8.77

A) (

1) O

R u

se P

=Ipe

akV p

eak/2

(1)

• U

se o

f t

VIE=

(1)

• t =

264

s (1

)

Exam

ple

of c

alcu

latio

n:

J10

5.33

kgJ10

2.37

kg225

0.5

16

×=

××

==

Δ−

Lm

E

V =

325

V/√

2 =

230

V a

nd I

= 12

.4 A

/√2

= 8.

77 A

s264

A8.75

V230

J10

5.33

5

=××

==VIE

t

4

(Tot

al fo

r Q

uest

ion

17 =

8 m

arks

)


93

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

18 (a

)*

This

que

stio

n as

sess

es a

stud

ent’s

abi

lity

to sh

ow a

coh

eren

t an

d lo

gica

lly st

ruct

ured

ans

wer

with

link

ages

and

fully

-su

stai

ned

reas

onin

g.

Mar

ks a

re a

war

ded

for i

ndic

ativ

e co

nten

t and

for h

ow th

e an

swer

is st

ruct

ured

and

show

s lin

es o

f rea

soni

ng.

The

follo

win

g ta

ble

show

s how

the

mar

ks sh

ould

be

awar

ded

for i

ndic

ativ

e co

nten

t. N

umbe

r of

indi

cativ

e m

arki

ng

poin

ts se

en in

an

swer

Num

ber o

f m

arks

aw

arde

d fo

r ind

icat

ive

mar

king

poi

nts

6 4

5–4

3 3–

2 2

1 1

0 0

Gui

danc

e on

how

the

mar

k sc

hem

e sh

ould

be

appl

ied:

The

mar

k fo

r ind

icat

ive

cont

ent s

houl

d be

add

ed to

th

e m

ark

for l

ines

of r

easo

ning

. Fo

r exa

mpl

e, a

n an

swer

with

five

indi

cativ

e m

arki

ng p

oint

s whi

ch is

pa

rtial

ly st

ruct

ured

with

som

e lin

kage

s and

line

s of

reas

onin

g sc

ores

4 m

arks

(3 m

arks

for i

ndic

ativ

e co

nten

t and

1 m

ark

for p

artia

l stru

ctur

e an

d so

me

linka

ges a

nd li

nes o

f rea

soni

ng).

If th

ere

are

no li

nkag

es b

etw

een

poin

ts, t

he sa

me

five

indi

cativ

e m

arki

ng p

oint

s wou

ld y

ield

an

over

all

scor

e of

3 m

arks

(3 m

arks

for i

ndic

ativ

e co

nten

t and

no

mar

ks fo

r lin

kage

s).

6


94

18 (a

)*

(con

tinue

d)

The

follo

win

g ta

ble

show

s how

the

mar

ks sh

ould

be

awar

ded

for s

truct

ure

and

lines

of r

easo

ning

.

N

umbe

r of m

arks

aw

arde

d fo

r stru

ctur

e of

ans

wer

and

su

stai

ned

line

of

reas

onin

g

Ans

wer

show

s a c

oher

ent a

nd

logi

cal s

truct

ure

with

link

ages

an

d fu

lly su

stai

ned

lines

of

reas

onin

g de

mon

stra

ted

thro

ugho

ut

2

Ans

wer

is p

artia

lly st

ruct

ured

w

ith so

me

linka

ges a

nd li

nes o

f re

ason

ing

1

Ans

wer

has

no

linka

ges

betw

een

poin

ts a

nd is

un

stru

ctur

ed

0


95

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

18 (a

)*

(con

tinue

d)

Indi

cativ

e co

nten

t •

a po

laris

ing

filte

r res

trict

s the

(ele

ctric

fiel

d) v

ibra

tions

of t

he

(tran

sver

se) l

ight

wav

e to

a si

ngle

pla

ne

• in

clud

ing

the

dire

ctio

n of

pro

paga

tion

of th

e lig

ht

• th

e lig

ht in

cide

nt o

n th

e fil

ter m

ust b

e pl

ane

pola

rised

•

whe

n th

e an

gle

of ro

tatio

n is

a m

ultip

le o

f π ra

d (in

clud

ing

zero

), th

e pl

ane

of p

olar

isat

ion

of th

e in

cide

nt li

ght i

s per

pend

icul

ar to

th

e tra

nsm

issi

on a

xis o

f the

pol

aris

ing

filte

r hen

ce th

e in

tens

ity

of th

e tra

nsm

itted

ligh

t is z

ero

• w

hen

the

angl

e of

rota

tion

is a

n od

d m

ultip

le o

f π/2

rad

the

plan

e of

pol

aris

atio

n of

the

inci

dent

ligh

t is t

he sa

me

as th

at o

f the

tra

nsm

issi

on a

xis o

f the

pol

aris

ing

filte

r hen

ce m

axim

um li

ght i

s tra

nsm

itted

•

the

inte

nsity

of t

he tr

ansm

itted

ligh

t var

ies f

rom

a m

inim

um to

a

max

imum

as t

he a

ngle

of r

otat

ion

varie

s as s

how

n by

the

grap

h


96

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

18 (a

)*

(con

tinue

d)

Alte

rnat

ive

answ

er

• a

pola

risin

g fil

ter r

estri

cts t

he (e

lect

ric fi

eld)

vib

ratio

ns o

f the

(tr

ansv

erse

) lig

ht w

ave

to a

sing

le d

irect

ion

• pe

rpen

dicu

lar t

o th

e di

rect

ion

of p

ropa

gatio

n of

the

light

•

the

light

inci

dent

on

the

filte

r is p

lane

pol

aris

ed

• w

hen

the

angl

e of

rota

tion

is a

mul

tiple

of π

rad

(incl

udin

g ze

ro),

the

plan

e of

pol

aris

atio

n of

the

inci

dent

ligh

t is p

erpe

ndic

ular

to

the

trans

mis

sion

axi

s of t

he p

olar

isin

g fil

ter h

ence

the

inte

nsity

of

the

trans

mitt

ed li

ght i

s zer

o •

whe

n th

e an

gle

of ro

tatio

n is

an

odd

mul

tiple

of π

/2 ra

d th

e pl

ane

of p

olar

isat

ion

of th

e in

cide

nt li

ght i

s the

sam

e as

that

of t

he

trans

mis

sion

axi

s of t

he p

olar

isin

g fil

ter h

ence

max

imum

ligh

t is

trans

mitt

ed

• th

e in

tens

ity o

f the

tran

smitt

ed li

ght v

arie

s fro

m a

min

imum

to a

m

axim

um a

s the

ang

le o

f rot

atio

n va

ries a

s sho

wn

by th

e gr

aph

18 (b

) •

Pass

ligh

t thr

ough

one

lens

of t

he g

lass

es a

nd v

iew

the

light

th

roug

h th

e le

ns o

f the

seco

nd p

air o

f gla

sses

. Rot

ate

one

pair

of

glas

ses t

hrou

gh 9

0° (1

) •

If th

e lig

ht in

tens

ity v

arie

s the

n th

e gl

asse

s use

pol

aris

ing

filte

rs

(1)

Allo

w fu

ll cr

edit

for a

suita

bly

anno

tate

d di

agra

m.

2 (T

otal

for

Que

stio

n 18

= 8

mar

ks)


97

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

19 (a

)(i)

An

expl

anat

ion

that

mak

es re

fere

nce

to:

• qu

antis

atio

n of

ene

rgy

is th

e id

ea th

at e

nerg

y is

em

itted

/radi

ated

in

dis

cret

e pa

cket

s/ph

oton

s (1)

•

each

pho

ton

has a

n en

ergy

whi

ch is

rela

ted

to fr

eque

ncy

OR

su

itabl

e re

fere

nce

to E

= h

f (1)

2

19 (a

)(ii)

•

Mod

el A

is su

cces

sful

at l

ong

wav

elen

gths

bec

ause

the

curv

e fo

r m

odel

A fo

llow

s the

exp

erim

enta

l cur

ve (1

) •

But

mod

el A

bre

aks d

own

for s

hort

wav

elen

gths

, sin

ce it

su

gges

ts th

at th

e in

tens

ity te

nds t

o in

finity

as t

he w

avel

engt

h ge

ts sh

orte

r (1)

•

Mod

el B

is su

cces

sful

for s

hort

wav

elen

gths

bec

ause

cur

ve B

fo

llow

s the

exp

erim

enta

l cur

ve (

1)

• B

ut m

odel

B in

dica

tes h

ighe

r tha

n ex

pect

ed in

tens

ities

at l

arge

r w

avel

engt

hs (1

)

4

19 (b

) •

Use

of c

=fλ

and

E=

hf (1

)

• C

onve

rts e

V to

J (1

)

• U

se o

f E =

W +

KE m

ax (1

)

• K

E max

= 1

.3 ×

10-1

9 J (1

)

Exam

ple

of c

alcu

latio

n:

J10

7.831

m10

2.54

sm

103

sJ10

6.63

197

18

34−

−

−−

×=

××

××

==λhc

E

Wor

k fu

nctio

n =

6.51

× 1

0-19

J

max

KEW

E+

=

J10

1.32

19J

-10

6.51

J10

7.83

-19

-19

max

×=

×−

×=

KE

4

(Tot

al fo

r Q

uest

ion

19 =

10

mar

ks)


98

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

20 (a

) •

The

wav

elen

gth

chan

ge is

big

ger t

he fu

rther

aw

ay th

e ga

laxi

es

are

(1)

• Th

e fu

rther

aw

ay g

alax

ies a

re th

e fa

ster

they

are

mov

ing,

so a

ll di

stan

t gal

axie

s are

mov

ing

away

from

eac

h ot

her (

and

the

univ

erse

is e

xpan

ding

) (1)

•

Ther

e is

a la

rge

amou

nt o

f sca

tter i

n H

ubbl

e’s o

rigin

al d

ata

set.

(1)

• Th

e or

igin

al d

ata

set c

over

s a v

ery

smal

l ran

ge o

f dis

tanc

es [o

nly

the

clos

est g

alax

ies c

onsi

dere

d] (1

) •

Hen

ce, o

n th

e ba

sis o

f the

orig

inal

dat

a, th

e co

nclu

sion

dra

wn

by

Hub

ble

was

qui

te sp

ecul

ativ

e (1

)

5

20 (b

) •

Use

of

3max

109.2

−×

=T

λ (1

)

• T

= 58

00 K

[acc

ept 5

780

K a

nd 6

000

K] (

1)

• U

se o

f 4

24

Tr

Lσ

π=

(1)

• r =

6.9×1

08 m (1

)

Exam

ple

of c

alcu

latio

n:

K5780

m10

5.02

mK

102.9

7-

-3

=×

×=

T

()

m10

6.91

5800K

KWm

105.67

4πW

103.85

84

42

8

26

×=

××

×

×=

−−

−r

4


99

Que

stio

n N

umbe

r

A

ccep

tabl

e an

swer

s

Add

ition

al g

uida

nce

Mar

k

20 (c

)(i)

• 8

alph

a de

cays

redu

ce th

e pr

oton

num

ber b

y 16

(1)

• pr

oton

num

ber d

ecre

ases

by

only

10,

so th

ere

mus

t be

6 β

deca

ys

(1)

OR

•

bala

nced

equ

atio

n w

ritte

n fo

r ove

rall

deca

y (1

) •

expl

icit

solu

tion

to g

ive

6 β

- dec

ays (

1)

Exam

ple

of c

alcu

latio

n:

−−×

+×

+→

βN

α8

PbU

0 14 2

206 82

238 92

92

= 8

2 +

(8×2

) – N

92

= 8

2 +

16 –

N

N =

98

- 92

= 6

Proo

f mus

t be

give

n to

obt

ain

thes

e m

arks

.

2 20

(c)(

ii)

• us

e of

2

ln2/1=

tλ

(1)

• us

e of

t

eN

Nλ

−=

0 (1

) •

N/N

0 = 0

.1 (1

) •

t = 2

.5 ×

105 y

ears

(1)

Exam

ple

of c

alcu

latio

n:

()

y10

2.49

y10

9.24

0.1

ln

ln

y10

9.24

75000

0.693

51

6

0

16

×=

×

−=

∴

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

−=

×=

=

−−

−−

t

NNt

λλ

4 (T

otal

for

Que

stio

n 20

= 1

5 m

arks

)

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1 (Total for Multiple Choice Questions = 10 marks) · 2015-09-23 · P v 3 1 2014 2014 81 2 P aper 2 Mark scheme Question Number Acceptable answers Additional guidance Mark 1 C 1