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A mesoatom of hydrogen is a hydrogen atom in which a negative muon with a mass 207 times that of an electron orbits the nucleus instead of an electron. Find the Bohr radii and the energy levels of a mesoatom. |
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Answer» `(m_(1)v_(1)^(2))/r_(1)=(m_(2)v_(2)^(2))/r_(2)=e^(2)/(4piepsi_(0)a^(2))` `a=r_(1)+r_(2)` `m_(1)v_(1)r_(1)+m_(2)v_(2)r_(2)=nh` `m_(1)r_(1)=m_(2)r_(2)` It follows from the first and the fourth equations that `v_(1)//r_(1)=v_(2)//r_(2)`, and from the third and the fourth that `v_(1)+v_(2)=(nh)/(m_(1)r_(1))=(nh)/(m_(2)r_(2))`  Hence we obtain the orbital speeds: `v_(1)=nh//am_(1),v_(2)=nh//am_(2)`. SUBSTITUTING this result into the first and the second equations, we obtain the radii: `r_(1)=n^(2)(4piepsi_(0)h^(2))/(m_(1)e^(2)),r_(2)=n^(2)(4piepsi_(0)h^(2))/(m_(2)e^(2))` Hence the Bohr radii of a mesoatom may be obtained: `a_(n)=n^(2)(4piepsi_(0)h^(2))/e^(2)*(m_(p)+m_(MU))/(m_(p)m_(mu))` The energy of an electron occupying an arbitrary energy level is `epsi_(n)=(m_(1)v_(1)^(2))/2+(m_(2)v_(2)^(2))/2-e^(2)/(4piepsi_(0)a_(n))=-e^(2)/(8piepsi_(0)a_(n))=-e^(2)/(8piepsi_(0)a_(0))*a_(0)/a_(n)` = `-hcRa_(0)/a_(n)` where `a_(0)` is the first Bohr radius of the hydrogen ATOM. |
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