# Photoelectric effect help

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Hello, I have been given the wavelength of a light source: 367nm and then the work function of a metal: 220KJmol-1. I have been asked to calculate whether the light source would release an electron. How do i do this?

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#2

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Hello, I have been given the wavelength of a light source: 367nm and then the work function of a metal: 220KJmol-1. I have been asked to calculate whether the light source would release an electron. How do i do this?

**Peace&Love**)Hello, I have been given the wavelength of a light source: 367nm and then the work function of a metal: 220KJmol-1. I have been asked to calculate whether the light source would release an electron. How do i do this?

Because the work function is the minimum energy required for the emission of the electrons of the metal surface, the kinetic energy of the ejected electrons is given by the equation K.E=hf-w where hf is the energy of the illuminated electromagnetic radiation and w the work function. But since f=c/lambda, the above equation becomes K.E=hc/(lambda) - w

For the emission of the electrons to occur, the energy of the radiation must

**at least**be equal to the work function of the metal, at which point the K.E of the electrons is zero but the energy of the radiation is

**just**equal to the work function.

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(Original post by

It's strange that you've been given the work function of the metal dependent on the amount of the metal because of its units (Jmol-1). The work function is independent of the amount of the metal.

Because the work function is the minimum energy required for the emission of the electrons of the metal surface, the kinetic energy of the ejected electrons is given by the equation K.E=hf-w where hf is the energy of the illuminated electromagnetic radiation and w the work function. But since f=c/lambda, the above equation becomes K.E=hc/(lambda) - w

For the emission of the electrons to occur, the energy of the radiation must

**Mehrdad jafari**)It's strange that you've been given the work function of the metal dependent on the amount of the metal because of its units (Jmol-1). The work function is independent of the amount of the metal.

Because the work function is the minimum energy required for the emission of the electrons of the metal surface, the kinetic energy of the ejected electrons is given by the equation K.E=hf-w where hf is the energy of the illuminated electromagnetic radiation and w the work function. But since f=c/lambda, the above equation becomes K.E=hc/(lambda) - w

For the emission of the electrons to occur, the energy of the radiation must

**at least**be equal to the work function of the metal, at which point the K.E of the electrons is zero but the energy of the radiation is**just**equal to the work function.yes I figured I had to use that equation but unsure how to use it given the work function I have been used. How do I go about converting the work function?

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#4

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yes I figured I had to use that equation but unsure how to use it given the work function I have been used. How do I go about converting the work function?

**Peace&Love**)yes I figured I had to use that equation but unsure how to use it given the work function I have been used. How do I go about converting the work function?

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#5

The work function is given per mole; think of this as the energy of a mole of electrons - divide by Avogadro's number, then convert to the corresponding wavelength of light (I reckon the light source would cause an electron to be emitted, by the way).

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#6

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The work function is given per mole; think of this as the energy of a mole of electrons - divide by Avogadro's number, then convert to the corresponding wavelength of light (I reckon the light source would cause an electron to be emitted, by the way).

**KombatWombat**)The work function is given per mole; think of this as the energy of a mole of electrons - divide by Avogadro's number, then convert to the corresponding wavelength of light (I reckon the light source would cause an electron to be emitted, by the way).

However, since a negatively charged electron on the metal surface is attracted to the positive nucleus, then a minimum amount of energy is required to overcome this electromagnetic force of attraction to remove the electron. This particular amount of energy is the work function of that specific metal. As a result, whether there is one atom or 1kg of the metal, the energy required to remove an outermost electron of the atom/metal is constant (because there is an individual interaction). If the work function of a metal was given per mole of the metal, then it would mean that the work function of the metal increases as the number of moles of the metal increases. Of course, when the energy of the radiation exceeds the work function of the metal, the energy required to remove all the electrons on the metal surface increases as the number of moles of the metal increases, but that's not the same as the work function of the metal.

Lets's conduct a mathematical approach to this: Since the work function, w, of a metal is defined per electron, then the work function of the whole metal is nw where n is the number of outermost electrons on the metal surface. But the energy of a photon not removing an electron but its energy being equal to the work function is hf

_{0}and so the energy of the whole radiation transferred per second is nhf

_{0}where, again, n is the number of photons emitted per second. Therefore the work function of the metal would be nhf

_{0}=nw or hf

_{0}=w

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#7

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Because the light is considered to contain photons each carrying an energy equal to hf, the simple assumption is that during the interaction of light with the metal surface each photon transfers its entire energy to a single electron (there is an individual interaction between each photon of light with an electron).

However, since a negatively charged electron on the metal surface is attracted to the positive nucleus, then a minimum amount of energy is required to overcome this electromagnetic force of attraction to remove the electron. This particular amount of energy is the work function of that specific metal. As a result, whether there is one atom or 1kg of the metal, the energy required to remove an outermost electron of the atom/metal is constant (because there is an individual interaction). If the work function of a metal was given per mole of the metal, then it would mean that the work function of the metal increases as the number of moles of the metal increases. Of course, when the energy of the radiation exceeds the work function of the metal, the energy required to remove all the electrons on the metal surface increases as the number of moles of the metal increases, but that's not the same as the work function of the metal.

Lets's conduct a mathematical approach to this: Since the work function, w, of a metal is defined per electron, then the work function of the whole metal is nw where n is the number of outermost electrons on the metal surface. But the energy of a photon not removing an electron but its energy being equal to the work function is hf

**Mehrdad jafari**)Because the light is considered to contain photons each carrying an energy equal to hf, the simple assumption is that during the interaction of light with the metal surface each photon transfers its entire energy to a single electron (there is an individual interaction between each photon of light with an electron).

However, since a negatively charged electron on the metal surface is attracted to the positive nucleus, then a minimum amount of energy is required to overcome this electromagnetic force of attraction to remove the electron. This particular amount of energy is the work function of that specific metal. As a result, whether there is one atom or 1kg of the metal, the energy required to remove an outermost electron of the atom/metal is constant (because there is an individual interaction). If the work function of a metal was given per mole of the metal, then it would mean that the work function of the metal increases as the number of moles of the metal increases. Of course, when the energy of the radiation exceeds the work function of the metal, the energy required to remove all the electrons on the metal surface increases as the number of moles of the metal increases, but that's not the same as the work function of the metal.

Lets's conduct a mathematical approach to this: Since the work function, w, of a metal is defined per electron, then the work function of the whole metal is nw where n is the number of outermost electrons on the metal surface. But the energy of a photon not removing an electron but its energy being equal to the work function is hf

_{0}and so the energy of the whole radiation transferred per second is nhf_{0}where, again, n is the number of photons emitted per second. Therefore the work function of the metal would be nhf_{0}=nw or hf_{0}=w
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#8

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Sure, you're right that the work function is intensive - but molar quantities are intensive too and you can definitely work out the energy of a mole of electrons. I know it's not very intuitive to think of a mole of electrons, but that's how the problems given, and chemists seem to like dealing with numbers in the hundreds of kJs (the numbers fit too, the energy given corresponds to photons of green light, if I've calculated it right).

**KombatWombat**)Sure, you're right that the work function is intensive - but molar quantities are intensive too and you can definitely work out the energy of a mole of electrons. I know it's not very intuitive to think of a mole of electrons, but that's how the problems given, and chemists seem to like dealing with numbers in the hundreds of kJs (the numbers fit too, the energy given corresponds to photons of green light, if I've calculated it right).

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#9

(Original post by

The general conclusion is the the work function is independent of the amount of the substance. Also, we are not given the moles of the substance and so if you've carried out the calculation, you've assumed the amount of the substance to be 1mole.

**Mehrdad jafari**)The general conclusion is the the work function is independent of the amount of the substance. Also, we are not given the moles of the substance and so if you've carried out the calculation, you've assumed the amount of the substance to be 1mole.

*actually*have. So the enthalpy of formation of, say, water is different if you had 1 g or 10 kg, but the

*molar*enthalpy of formation is always the same.

The same is true of the work function - you're right that it is normally given per electron, so is the same regardless of how many electrons are emitted. In this case we are given it per mole of electrons - the molar work function is the same regardless of how much metal you have, it is the energy of 1 mole of

**electrons**(i.e. 6.023*10^23 * (energy of 1 electron).

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#10

(Original post by

I should have said - an intensive property is one which doesn't change based on the amount of stuff you have. A molar property is always per mole of stuff, regardless of how many moles you

The same is true of the work function - you're right that it is normally given per electron, so is the same regardless of how many electrons are emitted. In this case we are given it per mole of electrons - the molar work function is the same regardless of how much metal you have, it is the energy of 1 mole of

**KombatWombat**)I should have said - an intensive property is one which doesn't change based on the amount of stuff you have. A molar property is always per mole of stuff, regardless of how many moles you

*actually*have. So the enthalpy of formation of, say, water is different if you had 1 g or 10 kg, but the*molar*enthalpy of formation is always the same.The same is true of the work function - you're right that it is normally given per electron, so is the same regardless of how many electrons are emitted. In this case we are given it per mole of electrons - the molar work function is the same regardless of how much metal you have, it is the energy of 1 mole of

**electrons**(i.e. 6.023*10^23 * (energy of 1 electron).@Peace&Love : now for simplicity, as the above user suggested previously, you need to divide the work function of the metal by Avogadro's constant so that you have the work function per electron. This is because the energy of the radiation calculated using hc/lamda is the energy of a photon.

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#11

**Peace&Love**)

Hello, I have been given the wavelength of a light source: 367nm and then the work function of a metal: 220KJmol-1. I have been asked to calculate whether the light source would release an electron. How do i do this?

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#12

(Original post by

work function is just plain energy?

**pinkfloydfan**)work function is just plain energy?

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