Higher Homework Physics Exercise 7- Wave – Particle Duality

Resources & Mark scheme

Resources:

Questions: / Past papers
1 / Q17 – 2002
2 / Q15-2004
3 / Q17 – 2006
4 / Q16- 2004
5 / Q15 - 2012
6 / Q17-2008
7 / Q
8 / Q17 – 2007
9 / Q15 – 2010
10 / Q28- 2000
11 / Q28- 2002
12 / Q27- 2003

Mark Scheme:

Q1 – E

Q2 – D

Q3 – A

Q4 – D

Q5 – E

Q6 – B

Q7 –

Q8 – E

Q9 – B

Q10:

28.a.i. Photoelectric emission is the term used to describe the process by which

an electron bound in an atom can absorb enough energy from a single photon

to escape, or be emitted, from the atom.

a.ii. Threshold frequency

a.iii.As the intensity of the radiation is increased there are more incident

photons on the metal. Consequently, more electrons can absorb energy

from the incident radiation. This will result in more emitted electrons

which is consistent with the increased current.

28.b.i. Ephoton= hf

Ephoton= 6.63x10-34x9.0x1014

Ephoton= 5.967x10-19J

b.ii. Etotal= NEphoton

N = Etotal/Ephoton

N = 40.5x10-6/5.967x10-19

N = 6.79x1013

OR (If you like doing things the hard way.)

I = P/A

I = (Etotal/t)/A

I = (NEphoton/t)/A

=>N = AIt/Ephoton

=>N = (1.8x10-9x25x15x60)/ 5.967x10-19

=>N = 4.05x10-5/5.967x10-19

=>N = 6.79x1013

b.iii.The time taken for sunlight to erase the chip will be greater.

This is because only a proportion of the 25W/m2 from sunlight is

ultraviolet and it would therefore take longer for the semiconductor

to absorb the required number of photons.

Q11:

:

28.a. Threshold frequency.

28.b.i. f0 = 3.33x1014Hz

work function() = hf0

= 6.63x10-34x3.33x1014

= 2.21x10-19J

b.ii. Ephoton= hf

Ephoton= 6.63x10-34x 5.66x1014

Ephoton= 3.75x10-19J

Ek(electron) = Ephoton- 

Ek(electron) = 3.75x10-19 - 2.21x10-19

Ek(electron) = 1.54x10-19J

b.iii. Ek(gain) = qV

q = e = 1.6x10-19C

V = 2.00x104V

Ek(gain) = 1.6x10-19 x 2.00x104

Ek(gain) = 3.2x10-15J

Q12:

END OF EXERCISE 7