Фотоэффект құбылысы. Физика, 9 сынып, дидактикалық материал.

12.1 Photoelectric questions

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.

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

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 =  + eV_s, so  = hf - eV_{s, so}  = 1.07 x 10^-18 – (1.6 x 10^-19 x 3) = 5.9 x 10^-19 J

(c) eV_s = (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) eV_s = (1/2) mv² so eV_s = 1.00 x 10^-18 and V_s = 0.63 V