A photon at 6 × 10¹⁴ Hz has energy about 4 × 10⁻¹⁹ J and a 2 eV photon comes out as 3.2 × 10⁻¹⁹ J, so this IB sheet opens with two photon energy conversions. The answers record 4.8 × 10¹⁴ Hz and 6.63 × 10⁻¹⁰ m.
Key Takeaways
A 6 × 10¹⁴ Hz photon has energy about 4 × 10⁻¹⁹ J (Q1).
A 2 eV photon is 3.2 × 10⁻¹⁹ J (Q2).
Q4 records 4.8 × 10¹⁴ Hz as a frequency answer.
Section A: Multiple Choice Questions (1 Mark Each)
Choose the correct option for each question.
1.A photon has a frequency of 6 × 10¹⁴ Hz. Using h = 6.63 × 10⁻³⁴ J·s, its energy is approximately
3.In the photoelectric effect, increasing the intensity of the incident light increases
(a) the maximum kinetic energy of the photoelectrons(b) the number of photoelectrons emitted per second(c) the work function of the metal(d) the threshold frequency
4.A metal has a work function of 3.2 × 10⁻¹⁹ J. Using h = 6.63 × 10⁻³⁴ J·s, its threshold frequency is approximately
6.Which observation provides evidence for the wave nature of matter?
(a) the photoelectric effect(b) electron diffraction(c) black-body radiation(d) line emission spectra
Section B: Short Answer Type Questions (2 Marks Each)
Show all steps clearly.
7.Define a photon.
8.State the equation that relates the energy of a photon to its frequency.
9.Define the work function of a metal.
10.Explain what is meant by the threshold frequency in the photoelectric effect.
11.State how the maximum kinetic energy of photoelectrons depends on the frequency of the incident light.
12.State the de Broglie hypothesis.
13.State the Heisenberg uncertainty principle in words.
14.Explain why light below the threshold frequency produces no photoelectrons even at high intensity.
Section C: Numericals & Word Problems (3 Marks Each)
Apply the concepts to solve the problems. Show all working.
15.Using E = hf with h = 6.63 × 10⁻³⁴ J·s, calculate the energy of a photon with a frequency of 5 × 10¹⁴ Hz.
16.A photon has a wavelength of 500 nm. Using f = c ÷ λ and E = hf with c = 3 × 10⁸ m/s and h = 6.63 × 10⁻³⁴ J·s, calculate its frequency and energy.
17.A metal has a work function of 3.0 eV. Light of frequency 1.0 × 10¹⁵ Hz is incident on it. Using KE_max = hf − W with h = 6.63 × 10⁻³⁴ J·s and 1 eV = 1.6 × 10⁻¹⁹ J, calculate the maximum kinetic energy of the emitted electrons.
18.Photons of energy 5.0 eV strike a metal of work function 2.0 eV. Using KE_max = E_photon − W, calculate the maximum kinetic energy of the emitted electrons in eV and in joules.
19.Using p = mv and λ = h ÷ p, calculate the momentum and the de Broglie wavelength of an electron of mass 9.1 × 10⁻³¹ kg moving at 2 × 10⁶ m/s. Use h = 6.63 × 10⁻³⁴ J·s.
20.Using E = hf with h = 6.63 × 10⁻³⁴ J·s and 1 eV = 1.6 × 10⁻¹⁹ J, calculate the energy in eV of a photon of frequency 2.4 × 10¹⁴ Hz.
Answer Key
1. a) 4 × 10⁻¹⁹ J
2. a) 3.2 × 10⁻¹⁹ J
3. b) the number of photoelectrons emitted per second
4. a) 4.8 × 10¹⁴ Hz
5. a) 6.63 × 10⁻¹⁰ m
6. b) electron diffraction
7. Refer to solution guide.
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End of Worksheet
People Also Ask
What is the energy of a 6 × 10¹⁴ Hz photon?
About 4 × 10⁻¹⁹ J, option (a) in this worksheet (Q1).
How many joules is a 2 eV photon?
3.2 × 10⁻¹⁹ J, option (a) in this worksheet (Q2).
What does the photoelectric current depend on?
The number of photoelectrons emitted per second, option (b) in this worksheet (Q3).
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