Explaining the Photoelectric Effect lesson illustration


What is the relationship between E and f?

Explaining the Photoelectric Effect lesson illustration

 

Fill in the Blank Notes

 

II. Explaining photoelectric effect using quantum theory

 

Explaining the Photoelectric Effect lesson illustration

 

A. Quantum Theory

 

1) electrons can only absorb or emit energy in discrete amounts called quanta (packets)

 

 

2) Energy of each quanta (photon) is proportional to frequency of radiation

 

Explaining the Photoelectric Effect lesson illustration

 

Energy of a Quanta

 

 

E = hf

 

 

Explaining the Photoelectric Effect lesson illustration

 

June 2004 #40

 

Explaining the Photoelectric Effect lesson illustration
Max Planck

h = Planck's constant 

=6.63 X 10-34 Js

 

 

 

Since: c = fλ

c = 3 x 108 m/s

 

And f = c/λ

 


E = hc/λ


 

 

Check Your Understanding

5 quick questions — pick an answer to see if you’re right.

1. In quantum theory, electrons absorb or emit energy in discrete packets called…

2. The energy of a photon is…

3. Find the energy of a photon with frequency 1.0 × 10¹⁵ Hz. (h = 6.63 × 10⁻³⁴ J·s)

4. If a photon's frequency is doubled, its energy…

5. Light A has a longer wavelength than light B. Which photon has more energy?

NYSSLS Regents Practice · Modern Physics · Comparing Models

Wave Predictions vs. Observations

Scientists compared what the classical wave model predicted about the photoelectric effect with what experiments actually showed.

Table 1: Predictions vs. Observations

Variable ChangedWave Model PredictionObserved Result
Brighter lightElectrons gain more energyMore electrons; same max KE
Higher frequencyNo effect on energyMax KE increases
Very dim lightDelay before emissionEmission is immediate
Low-frequency lightEmission if bright enoughNo emission below a threshold

Which conclusion is best supported by Table 1?

▶ Want a challenge? Open 6 more Regents-style questions  (harder · mixed formats)

Challenge score: 0 / 6

Select all that apply

C1. Place a check mark in three boxes to indicate ideas of the photon model that explain the observed results in Table 1.

Multiple choice

C2. Why is emission immediate even in very dim light?

Substitute values

C3. Violet light has a frequency of 7.5 × 1014 Hz. Substitute numerical values into a mathematical representation that can be used to determine the energy of one violet photon in electronvolts. Then calculate the energy. (h = 6.63 × 10−34 J·s; 1 eV = 1.60 × 10−19 J)

E = ( × 10−34 J·s)( × 1014 Hz)(1.60 × 10−19 J/eV) = eV
Support or Refute · Justify

C4. A student makes the following claim.

Claim: Doubling the frequency of the light shining on a metal doubles the maximum kinetic energy of the ejected electrons.

Place a check mark in either the Support or Refute box. Justify your response using a mathematical relationship.

Multiple choice

C5. What happens if light below the threshold frequency is made 100 times brighter?

Constructed response

C6. Describe an investigation to test the “Higher frequency” row of Table 1. Identify the independent variable, the dependent variable, and one variable that must be kept constant, and state the result that would support the photon model.