Consider an hydrogen-like atom with a very light particle of mass m rotating around a heavy particle of mass M which is much greater than m. The charge of the heavy particle is -e and the charge of the lighter particle is +3e. The light particle executes uniform circular motion about its center of rotation which is where the heavy particle is located. Assume that the centripetal acceleration of the lighter particle is due essentially only to the Coulomb force of attraction between the two particles. Given that the orbital angular momentum of the light particle around its center of rotation is L = mur = ħ, where v is the lighter particle's speed and r is its distance from the center of rotation, determine the speed of the lighter particle relative to the heavy particle. HINT: you might want to make use of the relation ² = 137. ke² he Recall that: ħ = 1.06 x 10-34 J-s; k=1=9x 109 Nm²/C² and e = 1.6 x 10-19 C. 4reo
Consider an hydrogen-like atom with a very light particle of mass m rotating around a heavy particle of mass M which is much greater than m. The charge of the heavy particle is -e and the charge of the lighter particle is +3e. The light particle executes uniform circular motion about its center of rotation which is where the heavy particle is located. Assume that the centripetal acceleration of the lighter particle is due essentially only to the Coulomb force of attraction between the two particles. Given that the orbital angular momentum of the light particle around its center of rotation is L = mur = ħ, where v is the lighter particle's speed and r is its distance from the center of rotation, determine the speed of the lighter particle relative to the heavy particle. HINT: you might want to make use of the relation ² = 137. ke² he Recall that: ħ = 1.06 x 10-34 J-s; k=1=9x 109 Nm²/C² and e = 1.6 x 10-19 C. 4reo
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Transcribed Image Text:Consider a hydrogen-like atom with a very light particle of mass \( m \) rotating around a heavy particle of mass \( M \) which is much greater than \( m \). The charge of the heavy particle is \(-\frac{1}{2} e\) and the charge of the lighter particle is \(+3e\). The light particle executes uniform circular motion about its center of rotation which is where the heavy particle is located. Assume that the centripetal acceleration of the lighter particle is due essentially only to the Coulomb force of attraction between the two particles.
Given that the orbital angular momentum of the light particle around its center of rotation is \( L = mvr = \frac{1}{2} \hbar \), where \( v \) is the lighter particle's speed and \( r \) is its distance from the center of rotation, determine the speed of the lighter particle relative to the heavy particle.
HINT: you might want to make use of the relation \( \frac{k e^2}{hc} \approx \frac{1}{137} \).
Recall that: \( \hbar = 1.06 \times 10^{-34} \, \text{J} \cdot \text{s}; \, k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) and \( e = 1.6 \times 10^{-19} \, \text{C} \).
Expert Solution
Step 1
Given that
Mass of lighter particle=m
Mass of heavy particle= M
Charge on heavy particle=-e/3
Charge on lighter particle=+3e
Lighter particle executes uniform circular motion about it's center of rotation which is where heavy particle is located.
Angular momentum
mvr = ħ/2
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