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The longest wavelength in the Balmer series of hydrogen atom is given as 656 nm positronium can be considered as a system of an electron and a positron in which both these particles orbit each other. What will be the longest wavelength in the lyman series for positronium?
- a)121 nm
- b)156 nm
- c)243 nm
- d)276 nm
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The longest wavelength in the Balmer series of hydrogen atom is given ...
The expression of wavelength emitted when an electron makes a transition from nth level to mth level is given by
for Balmer series m = 2
The value for n for longest wavelength will be 3
for positronium mass m will be replaced with reduced mass μ.
Rydberg constant for H - atom
Rydberg constant for positronium
for lyman series
for longest wavelength n = 2
for Balmer series m = 2
The value for n for longest wavelength will be 3
for positronium mass m will be replaced with reduced mass μ.
Rydberg constant for H - atom
Rydberg constant for positronium
for lyman series
for longest wavelength n = 2
Most Upvoted Answer
The longest wavelength in the Balmer series of hydrogen atom is given ...
Understanding Positronium
Positronium is a unique atom-like structure formed by an electron and its antiparticle, the positron. Its behavior is similar to hydrogen, but due to the reduced mass of the system, the energy levels differ.
Key Concept: Energy Levels
- The energy levels of positronium can be calculated using the formula similar to hydrogen:
- E_n = - (μ * Z^2 * e^4) / (2 * n^2 * h^2)
- Here, μ is the reduced mass, Z is the atomic number, and n is the principal quantum number.
Reduced Mass of Positronium
- The reduced mass (μ) of positronium is given by:
- μ = (m_e * m_p) / (m_e + m_p)
- Since m_e = m_p (mass of electron and positron are equal), we have:
- μ = m_e / 2
Comparison with Hydrogen
- In hydrogen, the longest wavelength in the Lyman series (n=2 to n=1 transition) corresponds to 121 nm.
- The wavelength (λ) is inversely proportional to the energy difference between the levels.
Calculating the Wavelength for Positronium
- Since the reduced mass of positronium is half that of hydrogen, the energy levels will be halved.
- Therefore, the wavelength for the transition from n=2 to n=1 in positronium will be twice that of hydrogen:
- λ_pos = 2 * λ_Hydrogen = 2 * 121 nm = 242 nm.
Conclusion
- The longest wavelength in the Lyman series for positronium is approximately 243 nm (option C).
- This demonstrates the distinct behavior of positronium compared to hydrogen due to its unique particle composition.
Positronium is a unique atom-like structure formed by an electron and its antiparticle, the positron. Its behavior is similar to hydrogen, but due to the reduced mass of the system, the energy levels differ.
Key Concept: Energy Levels
- The energy levels of positronium can be calculated using the formula similar to hydrogen:
- E_n = - (μ * Z^2 * e^4) / (2 * n^2 * h^2)
- Here, μ is the reduced mass, Z is the atomic number, and n is the principal quantum number.
Reduced Mass of Positronium
- The reduced mass (μ) of positronium is given by:
- μ = (m_e * m_p) / (m_e + m_p)
- Since m_e = m_p (mass of electron and positron are equal), we have:
- μ = m_e / 2
Comparison with Hydrogen
- In hydrogen, the longest wavelength in the Lyman series (n=2 to n=1 transition) corresponds to 121 nm.
- The wavelength (λ) is inversely proportional to the energy difference between the levels.
Calculating the Wavelength for Positronium
- Since the reduced mass of positronium is half that of hydrogen, the energy levels will be halved.
- Therefore, the wavelength for the transition from n=2 to n=1 in positronium will be twice that of hydrogen:
- λ_pos = 2 * λ_Hydrogen = 2 * 121 nm = 242 nm.
Conclusion
- The longest wavelength in the Lyman series for positronium is approximately 243 nm (option C).
- This demonstrates the distinct behavior of positronium compared to hydrogen due to its unique particle composition.
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Community Answer
The longest wavelength in the Balmer series of hydrogen atom is given ...
The expression of wavelength emitted when an electron makes a transition from nth level to mth level is given by
for Balmer series m = 2
The value for n for longest wavelength will be 3
for positronium mass m will be replaced with reduced mass μ.
Rydberg constant for H - atom
Rydberg constant for positronium
for lyman series
for longest wavelength n = 2
for Balmer series m = 2
The value for n for longest wavelength will be 3
for positronium mass m will be replaced with reduced mass μ.
Rydberg constant for H - atom
Rydberg constant for positronium
for lyman series
for longest wavelength n = 2
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