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Before the neutrino hypothesis the beta ...

Before the neutrino hypothesis the beta decay process was throught to be the transition.
`n to p+bar(e)`
If this was true show that if the neutron was at rest the proton and electron would emerge with fixed energies and calculate them. Experimentally the electron energy was found to have a large range.

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To solve the problem, we will analyze the beta decay process where a neutron decays into a proton and an electron (beta particle). We will show that if the neutron is at rest, the proton and electron would emerge with fixed energies and calculate those energies. ### Step-by-Step Solution: 1. **Understanding the Initial Conditions**: - The neutron is at rest, which means its initial momentum \( P_n \) is zero. - The rest mass energy of the neutron is given by \( E_n = m_n c^2 \), where \( m_n \) is the mass of the neutron and \( c \) is the speed of light. ...
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The beta - decay process , discovered around 1900 , is basically the decay of a neutron n . In the laboratory , a proton p and an electron e^(bar) are observed as the decay product of neutron. Therefore considering the decay of neutron as a two- body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant . But experimentally , it was observed that the electron kinetic energy has continuous spectrum Considering a three- body decay process , i.e. n rarr p + e^(bar) + bar nu _(e) , around 1930 , Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino (bar nu_(e)) to be massaless and possessing negligible energy , and the neutrino to be at rest , momentum and energy conservation principle are applied. From this calculation , the maximum kinetic energy of the electron is 0.8 xx 10^(6) eV The kinetic energy carried by the proton is only the recoil energy. What is the maximum energy of the anti-neutrino ?

The beta - decay process , discovered around 1900 , is basically the decay of a neutron n . In the laboratory , a proton p and an electron e^(bar) are observed as the decay product of neutron. Therefore considering the decay of neutron as a two- body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant . But experimentally , it was observed that the electron kinetic energy has continuous spectrum Considering a three- body decay process , i.e. n rarr p + e^(bar) + bar nu _(e) , around 1930 , Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino (bar nu_(e)) to be massaless and possessing negligible energy , and the neutrino to be at rest , momentum and energy conservation principle are applied. From this calculation , the maximum kinetic energy of the electron is 0.8 xx 10^(6) eV The kinetic energy carried by the proton is only the recoil energy. If the - neutrono had a mass of 3 eV// c^(2) (where c is the speed of light ) insend of zero mass , what should be the range of the kinectic energy K. of the electron ?

Consider the decay of a free neutron at rest: n top+e^(-) Show that the tow-body dacay of this type must necessarily give an electron of fixed energy and, therefore, cannot for the observed continous energy distribution in the beta -decay of a neutron or a nucleus.

A metal foil is at a certain distance from an isotropic point source that emits energy at the rate P. Let us assume the classical physics to be applicable. The incident light energy will be absorbed continuously and smoothly. The electrons present in the foil soak up the energy incident on them. For simplicity , we can assume that the energy incident on a circular path of the foil with radius 5xx10^(-11) m (about that of a typical atom ) is absorbed by a single electron. The electron absorbs sufficient energy to break through the binding forces and comes out from the foil. By knowing the work function, we can calculate the time taken by an electron to come out i.e., we can find out the time taken by photoelectric emission to start. As you will see in the following questions, the time decay comes out to be large, which is not practically observed. The time lag is very small. Apparently, the electron does not have to soak up energy . It absorbs energy all at once in a single photon electron interaction. If power of the source P is 1.5 W and distance of the foil from the source is 3.5 m, the energy received by an electron per second is

A metal foil is at a certain distance from an isotropic point source that emits energy at the rate P. Let us assume the classical physics to be applicable. The incident light energy will be absorbed continuously and smoothly. The electrons present in the foil soak up the energy incident on them. For simplicity , we can assume that the energy incident on a circular path of the foil with radius 5xx10^(-11) m (about that of a typical atom ) is absorbed by a single electron. The electron absorbs sufficient energy to break through the binding forces and comes out from the foil. By knowing the work function, we can calculate the time taken by an electron to come out i.e., we can find out the time taken by photoelectric emission to start. As you will see in the following questions, the time decay comes out to be large, which is not practically observed. The time lag is very small. Apparently, the electron does not have to soak up energy . It absorbs energy all at once in a single photon electron interaction. if work function of the metal of the foil is 2.2 eV, the time taken by electron to come out is nearly

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The only element in the hydrogen atom resides under ordinary condition on the first orbit .When energy is supplied the element move to hjgher energy ornbit depending on the lower of energy absioerbed .When this electron to may of the electron return to any of the lower orbits, it emit energy Lyman series is formed when the electron to the lowest orbit white Balmer series ids formed when the electron returns to the second orbit similar Paschen Brackett, and Pfund series are formed when electron return to the third fourth , and fifth arbit from highest energy orbits, respectively Maximum number of liner produced is equal when as electron jumps from nth level to ground level is equal to (n(n - 1))/(2) If teh electron comes back from the energy level having energy E_(2) to the energy level having energy E_(1) then the difference may be expresent in terms of energy of photon as E_(2) - E_(1) = Delta E, lambda = hc//Delta E Since h and c are constants Delta E coresponding to definite energy , thus , each transition from one energy level to unother will produce a light of definite wavelem=ngth .This isd actually observed as a line in the spectrum of hydrogen atom Wave number of line is given by the formula bar v = RZ^(2)((1)/(n_(1)^(2))- (1)/(n_(12)^(2))) Where R is a Rydherg constant Its a single isolated atom, an electrons make transition from fifth excited state is second thern maximum number of different type of photon observed is

The only element in the hydrogen atom resides under ordinary condition on the first orbit .When energy is supplied the element move to hjgher energy ornbit depending on the lower of energy absioerbed .When this electron to may of the electron return to any of the lower orbits, it emit energy Lyman series is formed when the electron to the lowest orbit white Balmer series ids formed when the electron returns to the second orbit similar Paschen Brackett, and Pfund series are formed when electron return to the third fourth , and fifth arbit from highest energy orbits, respectively Maximum number of liner produced is equal when as electron jumps from nth level to ground level is equal to (n(n - 1))/(2) If teh electron comes back from the energy level having energy E_(2) to the energy level having energy E_(1) then the difference may be expresent in terms of energy of photon as E_(2) - E_(1) = Delta E, lambda = hc//Delta E Since h and c are constants Delta E coresponding to definite energy , thus , each transition from one energy level to unother will produce a light of definite wavelem=ngth .This isd actually observed as a line in the spectrum of hydrogen atom Wave number of line is given by the formula bar v = RZ^(2)((1)/(n_(1)^(2))- (1)/(n_(12)^(2))) Where R is a Rydherg constant The wave number of electromagnetic radiation emitted during the transition of electron in between the two levels of Li^(2+) ion whose pricipal quantum numbner sum is 4 and difference is 2 is

The only element in the hydrogen atom resides under ordinary condition on the first orbit .When energy is supplied the element move to hjgher energy ornbit depending on the lower of energy absioerbed .When this electron to may of the electron return to any of the lower orbits, it emit energy Lyman series is formed when the electron to the lowest orbit white Balmer series ids formed when the electron returns to the second orbit similar Paschen Brackett, and Pfund series are formed when electron return to the third fourth , and fifth arbit from highest energy orbits, respectively Maximum number of liner produced is equal when as electron jumps from nth level to ground level is equal to (n(n - 1))/(2) If teh electron comes back from the energy level having energy E_(2) to the energy level having energy E_(1) then the difference may be expresent in terms of energy of photon as E_(2) - E_(1) = Delta E, lambda = hc//Delta E Since h and c are constants Delta E coresponding to definite energy , thus , each transition from one energy level to unother will produce a light of definite wavelem=ngth .This isd actually observed as a line in the spectrum of hydrogen atom Wave number of line is given by the formula bar v = RZ^(2)((1)/(n_(1)^(2))- (1)/(n_(12)^(2))) Where R is a Rydherg constant If the ionisation potential for hydrogen -like atom in a sample is 122.4 V then the series limit of the paschen series for this atom is

The only element in the hydrogen atom resides under ordinary condition on the first orbit .When energy is supplied the element move to hjgher energy ornbit depending on the lower of energy absioerbed .When this electron to may of the electron return to any of the lower orbits, it emit energy Lyman series is formed when the electron to the lowest orbit white Balmer series ids formed when the electron returns to the second orbit similar Paschen Brackett, and Pfund series are formed when electron return to the third fourth , and fifth arbit from highest energy orbits, respectively Maximum number of liner produced is equal when as electron jumps from nth level to ground level is equal to (n(n - 1))/(2) If teh electron comes back from the energy level having energy E_(2) to the energy level having energy E_(1) then the difference may be expresent in terms of energy of photon as E_(2) - E_(1) = Delta E, lambda = hc//Delta E Since h and c are constants Delta E coresponding to definite energy , thus , each transition from one energy level to unother will produce a light of definite wavelem=ngth .This isd actually observed as a line in the spectrum of hydrogen atom Wave number of line is given by the formula bar v = RZ^(2)((1)/(n_(1)^(2))- (1)/(n_(12)^(2))) Where R is a Rydherg constant The difference in the wavelength of the second line is Lyman series and last line of breaker series is a hydrogen sample is

NCERT EXEMPLAR ENGLISH-SEMICONDUCTOR ELECTRONICS: MATERIAL,DEVICES AND SIMPLE CIRCUIT-Multiplel choice questions
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  10. Why do stable nuclei never have more protons than neutrons ?

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