Answer
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Hint: When an electron makes a transition from an excited state to the ground state, it will be closer to the ion of the atom. The potential energy of the electron will be negative since it is attractive while the kinetic energy will be positive.
Complete step by step answer:
When an electron makes a transition from an excited state to the ground state, it will come closer to the ion. It can also be interpreted as the principal number of the electron $n$ will decrease.
Now, the potential energy of the electron will be due to the charges of the ion and the electron as
$V(r) = - \dfrac{{kZ{e^2}}}{r}$ where $Z$ is the charge of the ion, $r$ is the distance of the electron from the ion which depends on the state of the electron.
When the electron comes to form an excited state to a ground state, its radius will decrease since it moves closer to the atom. As a result, the potential energy will become more negative so we can say that it decreases. Physically, this process can be looked at as the electron being more strongly attracted to the ion and as a result having a more negative potential energy.
Now the kinetic energy of the electron depends on the velocity of the electron. The velocity of the electron depends on the radius as \[v \propto \dfrac{1}{r}\]. So the kinetic energy $K \propto {v^2} \propto \dfrac{1}{{{r^2}}}$. Hence if the radius decreases in the ground state, the kinetic energy will increase.
Since the increase in kinetic energy is inversely proportional to the square of the distance while the potential is inversely proportional to the distance directly, the decrease in potential energy will overpower the increase in kinetic energy. SO overall the total kinetic energy will also decrease.
So, option (A) is correct.
Note: We must be very careful in taking the potential energy as negative. Even if the magnitude of the potential energy increases, it will become more negative so overall it will decrease.
Complete step by step answer:
When an electron makes a transition from an excited state to the ground state, it will come closer to the ion. It can also be interpreted as the principal number of the electron $n$ will decrease.
Now, the potential energy of the electron will be due to the charges of the ion and the electron as
$V(r) = - \dfrac{{kZ{e^2}}}{r}$ where $Z$ is the charge of the ion, $r$ is the distance of the electron from the ion which depends on the state of the electron.
When the electron comes to form an excited state to a ground state, its radius will decrease since it moves closer to the atom. As a result, the potential energy will become more negative so we can say that it decreases. Physically, this process can be looked at as the electron being more strongly attracted to the ion and as a result having a more negative potential energy.
Now the kinetic energy of the electron depends on the velocity of the electron. The velocity of the electron depends on the radius as \[v \propto \dfrac{1}{r}\]. So the kinetic energy $K \propto {v^2} \propto \dfrac{1}{{{r^2}}}$. Hence if the radius decreases in the ground state, the kinetic energy will increase.
Since the increase in kinetic energy is inversely proportional to the square of the distance while the potential is inversely proportional to the distance directly, the decrease in potential energy will overpower the increase in kinetic energy. SO overall the total kinetic energy will also decrease.
So, option (A) is correct.
Note: We must be very careful in taking the potential energy as negative. Even if the magnitude of the potential energy increases, it will become more negative so overall it will decrease.
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