A neutron breaks into a proton and an electron. This decay of neutron is accompanied by the release of energy.Assuming that 50% of the energy is produced in the form of electromagnetic radiation, what will be the frequency of radiation produced.?

Answer 1

1st part
A free neutron decays forming particles such as proton #(p)# ,electron #(c^-)# and antineutrino #(barnu_e)# along with electromagnetic radiation in the form of #gamma#-ray

#n->p+e^(-)+ barnu_e+gamma#

Here the mass difference occurring for this transformation goes to the formation of electromagnetic radiation.

Initial mass.

the mass of neutron #m_n= 939.5656 MeV#

Final mass

the mass of proton #m_p= 938.2723 MeV#

the mass of electron #m_e=0.5100 MeV#

So mass difference (neglecting mass of antineutrino

#Deltam=m_n-m_p-m_e=0.7833MeV#

So total energy produced #DeltaE=0.7833MeV#

As per problem #50%# of total energy is produced in the form of electromagnetic radiation.

So radiated energy #DeltaE_"rad"=1/2DeltaE=0.3917MeV#

So frequency of radiation #nu=(DeltaE_"rad")/h#.

#=>nu=(0.3917xx10^6eV)/(4.1356xx10^-15eVs)=color(red)(9.4714xx10^19Hz)#.

where #h->"planck's constant"=4.1356xx10^-15eVs#

2nd part

Given 1st ionization energy of Al

#E_"ionisation"=577kJmol^(-1)#

#=577kJmol^(-1)xx1/(1.6022xx10^-22(kJ)/(eV))xx1/(6.022xx10^23mol^-1)#

#=5.98eV=5.98xx10^-6MeV#

So this energy is needed to remove first electron from the outermost shell of an atom of Al in its lowest energy state.

The energy of radiation #DeltaE_"rad"=0.3917MeV# being much higher than 1st ionization energy of Aluminum atom electron will be ejected absorbing the energy of photon of the given radiation,

The energy of the ejected electron

#E_"electron ejected"=DeltaE_"rad"-E_"ionisation"#

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Answer 2

The frequency of the radiation produced when a neutron breaks into a proton and an electron, with 50% of the energy released in the form of electromagnetic radiation, cannot be determined without knowing the exact energy released in the decay process. The frequency of the radiation is directly proportional to the energy of the radiation according to the equation E = hf, where E is the energy of the radiation, h is Planck's constant, and f is the frequency of the radiation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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