What happens to the first ionization potential when a hydrogen-like atom captures a particle?
$begingroup$
This is a textbook problem from Resonance DLPD Physical Chemistry, Page #83:
The mass of a proton is $1836$ times the mass of an electron. If a subatomic particle of mass $207$ times the mass of an electron is captured by the nucleus, what happens to the first ionization potential of H?
My answer is that it may either increase or decrease depending on the charge of the captured particle.
However, the correct answer according to my book is that the ionization potential increases.
How do I arrive at this solution?
physical-chemistry ionization-energy atomic-structure
asked 2 hours ago
user69284user69284
434
$endgroup$
add a comment |
$begingroup$
This is a textbook problem from Resonance DLPD Physical Chemistry, Page #83:
The mass of a proton is $1836$ times the mass of an electron. If a subatomic particle of mass $207$ times the mass of an electron is captured by the nucleus, what happens to the first ionization potential of H?
My answer is that it may either increase or decrease depending on the charge of the captured particle.
However, the correct answer according to my book is that the ionization potential increases.
How do I arrive at this solution?
physical-chemistry ionization-energy atomic-structure
asked 2 hours ago
user69284user69284
434
$endgroup$
1
$begingroup$
Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
$endgroup$
– Soumik Das
2 hours ago
add a comment |
$begingroup$
This is a textbook problem from Resonance DLPD Physical Chemistry, Page #83:
The mass of a proton is $1836$ times the mass of an electron. If a subatomic particle of mass $207$ times the mass of an electron is captured by the nucleus, what happens to the first ionization potential of H?
My answer is that it may either increase or decrease depending on the charge of the captured particle.
However, the correct answer according to my book is that the ionization potential increases.
How do I arrive at this solution?
physical-chemistry ionization-energy atomic-structure
asked 2 hours ago
user69284user69284
434
$endgroup$
This is a textbook problem from Resonance DLPD Physical Chemistry, Page #83:
The mass of a proton is $1836$ times the mass of an electron. If a subatomic particle of mass $207$ times the mass of an electron is captured by the nucleus, what happens to the first ionization potential of H?
My answer is that it may either increase or decrease depending on the charge of the captured particle.
However, the correct answer according to my book is that the ionization potential increases.
How do I arrive at this solution?
physical-chemistry ionization-energy atomic-structure
physical-chemistry ionization-energy atomic-structure
asked 2 hours ago
user69284user69284
434
asked 2 hours ago
user69284user69284
434
asked 2 hours ago
user69284user69284
434
asked 2 hours ago
user69284user69284
434
asked 2 hours ago
user69284user69284
434
434
1
$begingroup$
Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
$endgroup$
– Soumik Das
2 hours ago
add a comment |
1
$begingroup$
Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
$endgroup$
– Soumik Das
2 hours ago
1
1
$begingroup$
Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
$endgroup$
– Soumik Das
2 hours ago
$begingroup$
Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
$endgroup$
– Soumik Das
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:
$$ psi_{1s} = frac{2}{a_o^{3/2}} e^{-r/a_0} tag{1} $$
the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:
$$ a_0 = frac{hbar^2}{me^2} tag{2} $$
The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:
$$ m = frac{m_e m_p}{m_e + m_p} tag{3} $$
If we take the limit of $m_p to infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.
Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:
$$ I = frac{me^4}{8 epsilon_0 ^2 n^2h^2} tag{4} $$
the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.
A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?
answered 1 hour ago
John RennieJohn Rennie
1,422918
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:
$$ psi_{1s} = frac{2}{a_o^{3/2}} e^{-r/a_0} tag{1} $$
the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:
$$ a_0 = frac{hbar^2}{me^2} tag{2} $$
The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:
$$ m = frac{m_e m_p}{m_e + m_p} tag{3} $$
If we take the limit of $m_p to infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.
Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:
$$ I = frac{me^4}{8 epsilon_0 ^2 n^2h^2} tag{4} $$
the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.
A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?
answered 1 hour ago
John RennieJohn Rennie
1,422918
$endgroup$
add a comment |
$begingroup$
When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:
$$ psi_{1s} = frac{2}{a_o^{3/2}} e^{-r/a_0} tag{1} $$
the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:
$$ a_0 = frac{hbar^2}{me^2} tag{2} $$
The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:
$$ m = frac{m_e m_p}{m_e + m_p} tag{3} $$
If we take the limit of $m_p to infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.
Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:
$$ I = frac{me^4}{8 epsilon_0 ^2 n^2h^2} tag{4} $$
the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.
A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?
answered 1 hour ago
John RennieJohn Rennie
1,422918
$endgroup$
add a comment |
$begingroup$
When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:
$$ psi_{1s} = frac{2}{a_o^{3/2}} e^{-r/a_0} tag{1} $$
the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:
$$ a_0 = frac{hbar^2}{me^2} tag{2} $$
The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:
$$ m = frac{m_e m_p}{m_e + m_p} tag{3} $$
If we take the limit of $m_p to infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.
Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:
$$ I = frac{me^4}{8 epsilon_0 ^2 n^2h^2} tag{4} $$
the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.
A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?
answered 1 hour ago
John RennieJohn Rennie
1,422918
$endgroup$
When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:
$$ psi_{1s} = frac{2}{a_o^{3/2}} e^{-r/a_0} tag{1} $$
the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:
$$ a_0 = frac{hbar^2}{me^2} tag{2} $$
The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:
$$ m = frac{m_e m_p}{m_e + m_p} tag{3} $$
If we take the limit of $m_p to infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.
Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:
$$ I = frac{me^4}{8 epsilon_0 ^2 n^2h^2} tag{4} $$
the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.
A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?
answered 1 hour ago
John RennieJohn Rennie
1,422918
answered 1 hour ago
John RennieJohn Rennie
1,422918
answered 1 hour ago
John RennieJohn Rennie
1,422918
answered 1 hour ago
John RennieJohn Rennie
1,422918
1,422918
add a comment |
add a comment |
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$begingroup$
Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
$endgroup$
– Soumik Das
2 hours ago