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The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2 is :
- a)36x/5
- b)16x/7
- c)9x/5
- d)5x/9
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The shortest wavelength of He atom in Balmer series is x, then longest...
The shortest wavelength of He+ ion spectrum in Balmer series is the transition from n = infinity ----> n = 2.
Hence, 1/λ1 = 1/x = RZ^2(1/2^2 − 1/∞^2)= R 2^2(1/4) = R
or R = 1/x
Then, the longest wavelength in the Paschen series of line of Li2+ ion will be from n = 4 -----> n = 3
1/λ2 = RZ^2(1/3^2 − 1/4^2) = R 3^2(1/9 − 1/16) = R 9(7/144)=R 7/16
or Since, R = 1/x
1/λ2 = 1/x x 7/16
Or λ2 =16x/7
Most Upvoted Answer
The shortest wavelength of He atom in Balmer series is x, then longest...
Shortest wavelength of He atom in Balmer series
The Balmer series of hydrogen-like atoms consists of spectral lines that result from electron transitions from energy levels n ≥ 3 to n = 2. For He atom, the Balmer series corresponds to n = 3 to n = 2 transitions.
Let λ be the shortest wavelength in the Balmer series of He atom. Then, we have:
1/λ = R(1/2^2 - 1/3^2)
where R is the Rydberg constant for He atom.
Simplifying the above equation, we get:
λ = 8R/5
Longest wavelength in the Paschen series of Li 2
The Paschen series of hydrogen-like atoms consists of spectral lines that result from electron transitions from energy levels n ≥ 4 to n = 3. For Li 2, the Paschen series corresponds to n = 4 to n = 3 transitions.
Let λ' be the longest wavelength in the Paschen series of Li 2. Then, we have:
1/λ' = R'(1/3^2 - 1/4^2)
where R' is the Rydberg constant for Li 2.
Simplifying the above equation, we get:
λ' = 16R'/7
Relation between Rydberg constants for He and Li 2
The Rydberg constant for hydrogen-like atoms is given by:
R∝m/e^2
where m is the reduced mass of the atom and e is the charge on the electron.
For He atom, we have:
m = 2m_e
where m_e is the mass of electron.
For Li 2, we have:
m = m_e/2
Substituting these values in the above equation, we get:
R/R' = 4
Using this relation, we can write:
λ' = 16R'/7 = 16R/(4*7) = 4λ/7
Substituting the value of λ from the first part, we get:
λ' = 16x/7
Therefore, the correct option is B.
The Balmer series of hydrogen-like atoms consists of spectral lines that result from electron transitions from energy levels n ≥ 3 to n = 2. For He atom, the Balmer series corresponds to n = 3 to n = 2 transitions.
Let λ be the shortest wavelength in the Balmer series of He atom. Then, we have:
1/λ = R(1/2^2 - 1/3^2)
where R is the Rydberg constant for He atom.
Simplifying the above equation, we get:
λ = 8R/5
Longest wavelength in the Paschen series of Li 2
The Paschen series of hydrogen-like atoms consists of spectral lines that result from electron transitions from energy levels n ≥ 4 to n = 3. For Li 2, the Paschen series corresponds to n = 4 to n = 3 transitions.
Let λ' be the longest wavelength in the Paschen series of Li 2. Then, we have:
1/λ' = R'(1/3^2 - 1/4^2)
where R' is the Rydberg constant for Li 2.
Simplifying the above equation, we get:
λ' = 16R'/7
Relation between Rydberg constants for He and Li 2
The Rydberg constant for hydrogen-like atoms is given by:
R∝m/e^2
where m is the reduced mass of the atom and e is the charge on the electron.
For He atom, we have:
m = 2m_e
where m_e is the mass of electron.
For Li 2, we have:
m = m_e/2
Substituting these values in the above equation, we get:
R/R' = 4
Using this relation, we can write:
λ' = 16R'/7 = 16R/(4*7) = 4λ/7
Substituting the value of λ from the first part, we get:
λ' = 16x/7
Therefore, the correct option is B.
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The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer?
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