Kinetic Energy of the Photoelectron lesson illustration

 

Which color laser was able to make the paper glow the most? Why?

 

K.E. of Photoelectron
Was Determined By:

 

 

a) frequency of light (color)

Kinetic Energy of the Photoelectron lesson illustration     

 

 

 

photoelectric.mov

Rutgers University

 

 

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The white light was unable to discharge the electroscope  (eject the electrons) but the ultraviolet light could. Why???

The ultraviolet light had a higher frequency ....

... a higher energy.

 

 

b) nature of emitting metal

Typical Plot of Frequency of
Light vs. K.E. of Photoelectron

KE of
Photoelectron
Kinetic Energy of the Photoelectron lesson illustration

fo - Threshold Frequency - frequency below which no photoelectrons will be emitted

Kinetic Energy of the Photoelectron lesson illustration

Below Threshold?

 

 

No Emission

 

 

 

 

 Above Threshold?

 

 

 

 Emission

 

 

 

Kinetic Energy of the Photoelectron lesson illustration

 

Check Your Understanding

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

1. The kinetic energy of a photoelectron is determined by the light's frequency and…

2. What is the threshold frequency?

3. Ultraviolet light discharged an electroscope but white light did not. Why?

4. Very bright light below a metal's threshold frequency shines on it. What happens?

5. Which color of visible light would give photoelectrons the most kinetic energy from the same metal?

NYSSLS Regents Practice · Modern Physics · Energy of Photoelectrons

A Calculator’s Solar Cell Material

A metal with a work function of 2.3 eV is illuminated with light of wavelength 4.0 × 10−7 m (violet). (h = 6.63 × 10−34 J·s; c = 3.00 × 108 m/s; 1 eV = 1.60 × 10−19 J)

What is the maximum kinetic energy of the emitted electrons?

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

Challenge score: 0 / 6

Multiple choice

C1. Red light (7.0 × 10−7 m, 1.8 eV) shines on the same metal. What happens?

Determine the value

C2. Determine the longest wavelength of light that can eject electrons from this metal (work function 2.3 eV). Give your answer in nanometers (1 nm = 10−9 m).

λ0 = nm
Support or Refute · Justify

C3. A student makes the following claim.

Claim: If the wavelength of the light is halved to 2.0 × 10−7 m, the maximum kinetic energy of the electrons will double, from 0.8 eV to 1.6 eV.

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

Substitute values

C4. The fastest electrons have a kinetic energy of 1.3 × 10−19 J. Substitute numerical values into a mathematical representation that can be used to determine their speed. Then calculate the speed. (me = 9.11 × 10−31 kg)

v = √( 2( × 10−19 J)( × 10−31 kg) ) = × 105 m/s
Multiple choice

C5. Which change would increase the maximum kinetic energy of the photoelectrons?

Constructed response

C6. An engineer wants to use this metal (work function 2.3 eV) in a light meter for outdoor sunlight. Identify one benefit and one drawback of this choice, using the photoelectric equation.