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I dont understand what the mark scheme of q12 b means why cant thbeta radiation comes from electron in atomic nucleus? Why is it not good to have large energy and why energy is associated with wavelength?

Please help!

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I dont understand what the mark scheme of q12 b means why cant thbeta radiation comes from electron in atomic nucleus? Why is it not good to have large energy and why energy is associated with wavelength?

Please help!

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**Lamalam**)I dont understand what the mark scheme of q12 b means why cant thbeta radiation comes from electron in atomic nucleus? Why is it not good to have large energy and why energy is associated with wavelength?

Please help!

Posted from TSR Mobile

c = wavelength*frequency

Where c is the speed of light.

Since c is a constant, if wavelength is small, frequency is large.

Also,

Energy = frequency*plank's constant

Therefore if frequency is large, energy is also large.

Therefore, if wavelength is small, energy is large.

But you would expect the potential energy near the nucleus to be very low, as high energy orbits are further from the nucleus. Such a high energy electron near the nucleus just doesn't make sense, because you would think that with such a high energy it would be able to escape the nucleus.

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Energy and wavelength are related. As wavelength decreases, energy increases. This can be shown mathematically by the equation:

c = wavelength*frequency

Where c is the speed of light.

Since c is a constant, if wavelength is small, frequency is large.

Also,

Energy = frequency*plank's constant

Therefore if frequency is large, energy is also large.

Therefore, if wavelength is small, energy is large.

But you would expect the potential energy near the nucleus to be very low, as high energy orbits are further from the nucleus. Such a high energy electron near the nucleus just doesn't make sense, because you would think that with such a high energy it would be able to escape the nucleus.

**Doctor_Einstein**)Energy and wavelength are related. As wavelength decreases, energy increases. This can be shown mathematically by the equation:

c = wavelength*frequency

Where c is the speed of light.

Since c is a constant, if wavelength is small, frequency is large.

Also,

Energy = frequency*plank's constant

Therefore if frequency is large, energy is also large.

Therefore, if wavelength is small, energy is large.

But you would expect the potential energy near the nucleus to be very low, as high energy orbits are further from the nucleus. Such a high energy electron near the nucleus just doesn't make sense, because you would think that with such a high energy it would be able to escape the nucleus.

i have a confusion over the sue of equation like:

E=hf

c=f (lambda) (or v= f times lambda, when should I use c and when should i use v?)

Kinetic energy=(m*v^2)/2

einstein's photoelectric equation hf = Φ + E

_{kmax' which equation applies to photon only and which applies to electron only?}

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thank you

i have a confusion over the sue of equation like:

E=hf

c=f (lambda) (or v= f times lambda, when should I use c and when should i use v?)

Kinetic energy=(m*v^2)/2

einstein's photoelectric equation hf = Φ + E

**Lamalam**)thank you

i have a confusion over the sue of equation like:

E=hf

c=f (lambda) (or v= f times lambda, when should I use c and when should i use v?)

Kinetic energy=(m*v^2)/2

einstein's photoelectric equation hf = Φ + E

_{kmax' which equation applies to photon only and which applies to electron only?}For both electrons and photons, E = hf is true.

KE = 0.5mv^2 only applies for particles that have mass, like electrons, and doesn't apply to massless particles like photons.

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You're right that c = f*lambda only applies for light, not electrons, but for electrons v = f*lambda, where v is the phase velocity of the electron.

For both electrons and photons, E = hf is true.

KE = 0.5mv^2 only applies for particles that have mass, like electrons, and doesn't apply to massless particles like photons.

**Doctor_Einstein**)You're right that c = f*lambda only applies for light, not electrons, but for electrons v = f*lambda, where v is the phase velocity of the electron.

For both electrons and photons, E = hf is true.

KE = 0.5mv^2 only applies for particles that have mass, like electrons, and doesn't apply to massless particles like photons.

is einstein's photoelectric equation hf = Φ + E

_{kmax' for electron also? cause I only encounter the equation in photoelectric effect .}

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thanks!

is einstein's photoelectric equation hf = Φ + E

**Lamalam**)thanks!

is einstein's photoelectric equation hf = Φ + E

_{kmax' for electron also? cause I only encounter the equation in photoelectric effect .}Emax = hf - workfunction

In order to understand this equation you have to understand the process this equation describes. An incident light beam, with frequency f, collides with a metallic surface. The workfunction of the metallic surface is the energy required to ionize one electron from the metallic surface.

In this scenario, hf is the energy of the incoming light. Some of this energy is used to ionize the electron it collides with, and some of the energy is converted to kinetic energy in the ionized electron. The term (hf - workfunction) represents the left over energy after ionizing an electron on a metallic surface. This leftover energy is converted into kinetic energy in the electron.

Therefore Emax is the kinetic energy of the ejected electron, given by hf - workfunction, where f is the frequency of the incident light beam.

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