# Tap 502-2: Photoelectric effect questions

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TAP 502-2: Photoelectric effect questions
hf =  + (1/2) mv2 and hf =  + eVs
e = 1.60 x 10-19 C,

h = 6.63 x 10-34 J s,

mass of electron = 9.11 x 10-31 kg
1 The work function for lithium is 4.6 x 10-19 J.

(a) Calculate the lowest frequency of light that will cause photoelectric emission.

(b) What is the maximum energy of the electrons emitted when light of 7.3 x 1014 Hz is used?
2 Complete the table.

 Metal Work Function /eV Work Function /J Frequency used /Hz Maximum KE of Ejected electrons /J Sodium 2.28 6 x 1014 Potassium 3.68 x 10-19 0.32 x 10-19 Lithium 2.9 1 x 1015 Aluminium 4.1 0.35 x 10-19 Zinc 4.3 1.12 x 10-19 Copper 7.36 x 10-19 1 x 1015

3 The stopping potential when a frequency of 1.61 x 1015 Hz is shone on a metal is 3 V.

(a) What is energy transferred by each photon?

(b) Calculate the work function of the metal.

(c) What is the maximum speed of the ejected electrons?
4 Selenium has a work function of 5.11 eV. What frequency of light would just eject electrons? (The threshold frequency is when the max KE of the ejected electrons is zero)
5 A frequency of 2.4 x 1015 Hz is used on magnesium with work function of 3.7 eV.

(a) What is energy transferred by each photon?

(b) Calculate the maximum KE of the ejected electrons.

(c) The maximum speed of the electrons.

(d) The stopping potential for the electrons.

Answers and worked solutions

1(a) hf = 

hf = 4.60 x 10-19

f = 4.60 x 10-19 / 6.63 x 10-34 = 6.94 x 1014 Hz

(b) hf =  + (1/2) mv2. (6.63 x 10-34 x 7.30 x 1014) = 4.60 x 10-19 + (1/2) mv2

4.84 x 10-19 - 4.60 x 10-19 = (1/2) mv2 = 0.24 x 10-19 J

2

 Metal Work Function / eV Work Function / J Frequency used / Hz Maximum KE of ejected electrons / J Sodium 2.28 3.65 x 10-19 6 x 1014 0.35 x 10-19 Potassium 2.30 3.68 x 10-19 6 x 1014 0.32 x 10-19 Lithium 2.90 4.64 x 10-19 1. x 1015 1.99 x 10-19 Aluminium 4.10 6.56 x 10-19 1.04 x 1015 0.35 x 10-19 Zinc 4.30 6.88 x 10-19 1.2 x 1015 1.12 x 10-19 Copper 4.60 7.36 x 10-19 1 x 1015 0

For copper 1 x 1015 Hz is below the threshold frequency so no electrons are ejected.

3

(a) 1.07 x 10-18 J

(b) hf =  + eVs, so  = hf - eVs, so  = 1.07 x 10-18 – (1.6 x 10-19 x 3) = 5.9 x 10-19 J

(c) eVs = (1/2) mv2 so (1.60 x 10-19 x 3) = 0.5 x 9.11 x 10-31 x v2

so v2 =1.04 x 1012 and v = 1.02 x 106 m s-1
4 1.2 x 1015 Hz
5

(a) 1.6 x 10-18 J

(b) (1/2) mv2 = 1.x 10-18 J

(c) v2 = 1.1 x 1012 so v = 1.1 x 106 m s-1

(d) eVs = (1/2) mv2 so eVs = 1.00 x 10-18 and Vs = 0.63 V

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