The photoelectric effect is the emission of electrons from a metal when suitable-frequency light shines, and photon energy is E = hf, so this Year 12 sheet opens with the effect and its energy rule. The work function answer records the minimum energy to remove an electron.
Key Takeaways
The photoelectric effect ejects electrons when suitable light shines (Q1).
Photon energy is E = hf (Q2).
Q3 records the minimum energy to remove an electron.
Section A: Multiple Choice Questions (1 Mark Each)
Choose the correct option for each question.
1.The photoelectric effect is the
(a) emission of electrons from a metal when light of a suitable frequency shines on it(b) absorption of electrons by a metal(c) reflection of light(d) production of photons
2.The energy of a photon is given by
(a) E = hf(b) E = h ÷ f(c) E = f ÷ h(d) E = h × λ
3.The work function is the
(a) minimum energy needed to remove an electron from a metal(b) energy of the incident photon(c) maximum kinetic energy of an electron(d) rest energy of the metal
4.Increasing the intensity of the incident light at a fixed frequency increases
(a) the number of photoelectrons(b) the kinetic energy of the electrons(c) the work function(d) the threshold frequency
5.The threshold frequency is the
(a) minimum frequency of light that can cause emission(b) maximum frequency of light(c) frequency of visible light(d) frequency of sound waves
Section B: Short Answer Type Questions (2 Marks Each)
Show all steps clearly.
7.State the photoelectric effect.
8.Write the formula for the energy of a photon.
9.Define the work function.
10.What is the threshold frequency?
11.What happens to the maximum kinetic energy of the photoelectrons when the frequency of the light is increased?
12.What happens to the photocurrent when the intensity of the light is increased at fixed frequency?
13.Give one application of the photoelectric effect.
14.State the value of Planck's constant.
Section C: Numericals & Word Problems (3 Marks Each)
Apply the concepts to solve the problems. Show all working.
15.A photon has a frequency of 5 × 10¹⁴ Hz. Using E = hf with h = 6.63 × 10⁻³⁴ J s, find its energy.
16.The work function of a metal is 2 eV. Convert this to joules using 1 eV = 1.6 × 10⁻¹⁹ J.
17.A photon has an energy of 6 × 10⁻¹⁹ J. Using f = E ÷ h with h = 6.63 × 10⁻³⁴ J s, find its frequency.
18.Light of frequency 1 × 10¹⁵ Hz has photon energy hf = 6.63 × 10⁻¹⁹ J. If the work function is 4 × 10⁻¹⁹ J, use hf = Φ + KE to find the maximum kinetic energy of the photoelectrons.
19.Convert an energy of 4 eV to joules using 1 eV = 1.6 × 10⁻¹⁹ J.
20.A photon has a frequency of 1 × 10¹⁵ Hz. Using E = hf with h = 6.63 × 10⁻³⁴ J s, find its energy.
Answer Key
1. a) emission of electrons from a metal when light of a suitable frequency shines on it
2. a) E = hf
3. a) minimum energy needed to remove an electron from a metal
4. a) the number of photoelectrons
5. a) minimum frequency of light that can cause emission
6. a) hf = Φ + ½mv²
7. Refer to solution guide.
8. Refer to solution guide.
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11. Refer to solution guide.
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20. Refer to solution guide.
End of Worksheet
People Also Ask
What is the photoelectric effect?
The emission of electrons from a metal when suitable-frequency light shines, option (a) in this worksheet (Q1).
What is the energy of a photon?
E = hf, option (a) in this worksheet (Q2).
What is the work function?
The minimum energy needed to remove an electron from a metal, option (a) in this worksheet (Q3).
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