Define electronvolt .

The electronvolt (eV) is the energy gained by an electron travelling through a potential difference of one volt .

To convert an energy from electronvolts to joules, you .......... by 1.6 x 10 -19 .

To convert an energy from electronvolts to joules, you multiply by 1.6 x 10 -19 .

True or False? The electronvolt is an SI unit of energy.

False. The SI unit of energy is the joule . The electronvolt is a non-SI unit used because quantum-scale energies are typically much smaller than one joule.

Define the photoelectric effect .

The photoelectric effect is the phenomenon in which electrons are emitted from the surface of a metal upon the absorption of electromagnetic radiation .

Define threshold frequency .

The minimum frequency of incident electromagnetic radiation required to remove a photoelectron from the surface of a metal.

Each electron in a metal can absorb only a .......... photon, which is evidence that light is quantised.

Each electron in a metal can absorb only a single photon, which is evidence that light is quantised.

Why do alkali metals (e.g. sodium) have threshold frequencies in the visible light region, while transition metals (e.g. zinc) have threshold frequencies in the ultraviolet region?

In alkali metals , the attractive forces between surface electrons and positive metal ions are relatively weak In transition metals , these attractive forces are much stronger , so higher-energy (UV) photons are needed to release an electron

Define the work function , Φ.

The minimum energy required to release a photoelectron from the surface of a metal.

Define stopping potential , V s .

The potential difference required to stop photoelectron emission from occurring.

True or False? Increasing the intensity of incident radiation increases the stopping potential of a photoelectric surface.

False. Increasing intensity increases the number of photoelectrons emitted (the photoelectric current ), but the stopping potential stays the same because intensity does not affect the maximum kinetic energy of the photoelectrons.

On a graph of E k(max) against frequency f , the .......... represents the threshold frequency f 0 .

On a graph of E k(max) against frequency f , the x-intercept represents the threshold frequency f 0 .

Why do photoelectrons emitted from the surface of a metal have the maximum kinetic energy, compared with those released from deeper within the metal?

Surface electrons need the least energy to escape, so more of the absorbed photon energy remains as kinetic energy. Electrons from deeper in the metal lose some of the absorbed energy reaching the surface, leaving them with less kinetic energy.

What is the photoelectric current proportional to, and why?

The photoelectric current is proportional to the intensity of the incident radiation. This is because intensity is proportional to the number of photons striking the metal per second, and since each photoelectron absorbs only one photon, the number of photoelectrons emitted per second is also proportional to the number of photons.

Define photoelectric current .

A measure of the number of photoelectrons emitted from a metal surface per second .

True or False? Increasing the intensity of light incident on a metal surface increases the maximum kinetic energy of the emitted photoelectrons.

False. Each photoelectron absorbs only one photon , so maximum kinetic energy depends only on the frequency of the incident radiation, not its intensity. Increasing intensity increases the number of photoelectrons emitted instead.

A LevelPhysicsAQAFlashcardsParticles & RadiationThe Photoelectric Effect

The Photoelectric Effect (AQA A Level Physics): Flashcards

Exam code: 7408

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FrontThe Electronvolt
Define electronvolt.

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Cards in this collection (23)

Define electronvolt.
The electronvolt (eV) is the energy gained by an electron travelling through a potential difference of one volt.
What is the value of one electronvolt in joules?
$1 eV = 1.60 \times 10^{- 19} J$
To convert an energy from electronvolts to joules, you .......... by 1.6 x 10^-19.
To convert an energy from electronvolts to joules, you multiply by 1.6 x 10^-19.
An electron accelerates from rest through a potential difference V. Give the equation relating the energy gained (in eV) to its kinetic energy.
$e V = \frac{1}{2} m v^{2}$
Give the equation for the speed v of an electron accelerated from rest through a potential difference V.
$v = \sqrt{\frac{2 e V}{m}}$
True or False?
The electronvolt is an SI unit of energy.
False.
The SI unit of energy is the joule. The electronvolt is a non-SI unit used because quantum-scale energies are typically much smaller than one joule.
Give the equation that defines potential difference V in terms of energy transferred E and charge Q.
$V = \frac{E}{Q}$
Define the photoelectric effect.
The photoelectric effect is the phenomenon in which electrons are emitted from the surface of a metal upon the absorption of electromagnetic radiation.
Define threshold frequency.
The minimum frequency of incident electromagnetic radiation required to remove a photoelectron from the surface of a metal.
Each electron in a metal can absorb only a .......... photon, which is evidence that light is quantised.
Each electron in a metal can absorb only a single photon, which is evidence that light is quantised.
Why do alkali metals (e.g. sodium) have threshold frequencies in the visible light region, while transition metals (e.g. zinc) have threshold frequencies in the ultraviolet region?
- In alkali metals, the attractive forces between surface electrons and positive metal ions are relatively weak
- In transition metals, these attractive forces are much stronger, so higher-energy (UV) photons are needed to release an electron
Define the work function, Φ.
The minimum energy required to release a photoelectron from the surface of a metal.
Define stopping potential, V_s.
The potential difference required to stop photoelectron emission from occurring.
Give the equation linking the maximum kinetic energy E_k(max) of a photoelectron to the stopping potential V_s.
$E_{k (m a x)} = e V_{s}$
True or False?
Increasing the intensity of incident radiation increases the stopping potential of a photoelectric surface.
False.
Increasing intensity increases the number of photoelectrons emitted (the photoelectric current), but the stopping potential stays the same because intensity does not affect the maximum kinetic energy of the photoelectrons.
State the photoelectric equation relating photon energy, work function and maximum kinetic energy.
$h f = Φ + E_{k (m a x)}$
What condition must be satisfied for a photon to just cause photoelectric emission, at the threshold frequency f₀?
$h f_{0} = Φ$
On a graph of E_k(max) against frequency f, the .......... represents the threshold frequency f₀.
On a graph of E_k(max) against frequency f, the x-intercept represents the threshold frequency f₀.
On a graph of E_k(max) against frequency f, what does the gradient represent?
Planck's constant, h.
Why do photoelectrons emitted from the surface of a metal have the maximum kinetic energy, compared with those released from deeper within the metal?
Surface electrons need the least energy to escape, so more of the absorbed photon energy remains as kinetic energy. Electrons from deeper in the metal lose some of the absorbed energy reaching the surface, leaving them with less kinetic energy.
What is the photoelectric current proportional to, and why?
The photoelectric current is proportional to the intensity of the incident radiation. This is because intensity is proportional to the number of photons striking the metal per second, and since each photoelectron absorbs only one photon, the number of photoelectrons emitted per second is also proportional to the number of photons.
Define photoelectric current.
A measure of the number of photoelectrons emitted from a metal surface per second.
True or False?
Increasing the intensity of light incident on a metal surface increases the maximum kinetic energy of the emitted photoelectrons.
False.
Each photoelectron absorbs only one photon, so maximum kinetic energy depends only on the frequency of the incident radiation, not its intensity. Increasing intensity increases the number of photoelectrons emitted instead.

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