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The wavelength of the first line of Balmer series of hydrogen atom is l Å. The wavelength of this line of a double ionised lithium atom Z = 3) is
- a)l/3
- b)l/9
- c)l/8
- d)l/27
Correct answer is option 'B'. Can you explain this answer?
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The wavelength of the first line of Balmer series of hydrogen atom is ...
Explanation:
The Balmer series of hydrogen atom is given by:
1/λ = R [1/2^2 - 1/n^2]
where λ is the wavelength of the spectral line, R is Rydberg constant and n is an integer greater than 2.
For the first line of Balmer series of hydrogen atom, n = 3. Therefore,
1/λ = R [1/2^2 - 1/3^2]
=> λ = 656.3 nm
The wavelength of the first line of Balmer series of hydrogen atom is 656.3 nm.
Now, let's consider the double ionized lithium atom (Li2+), with atomic number Z = 3.
As the lithium atom has lost two electrons, its effective nuclear charge is +3.
The formula for the wavelength of spectral lines of any atom is given by:
1/λ = Z^2 Rh [1/ni^2 - 1/nf^2]
where Rh is the Rydberg constant for the atom, ni and nf are the initial and final energy levels of the electron, and Z is the atomic number of the atom.
We need to find the wavelength of the first line of Balmer series for the Li2+ ion.
For the first line of Balmer series, nf = 2 and ni = 3.
Substituting the values in the above formula, we get:
1/λ = 3^2 Rh [1/3^2 - 1/2^2]
=> 1/λ = 9/36 Rh
=> λ = 4/9 Rh
The wavelength of the first line of Balmer series for the Li2+ ion is 4/9 times the wavelength of the first line of Balmer series for hydrogen.
Substituting the value of Rh for lithium (Rl) and Rh for hydrogen (R), we get:
λ = (4/9) (Rl/R) λ(H)
=> λ = (4/9) (Rl/R) (656.3 nm)
=> λ = (4/9) (4/9) λ(H)
=> λ = (16/81) λ(H)
=> λ = 656.3/9
=> λ = 72.92 nm
Therefore, the wavelength of the first line of Balmer series for the Li2+ ion is 72.92 nm, which is equal to λ(H)/9.
Hence, the correct answer is option (B) l/9.
The Balmer series of hydrogen atom is given by:
1/λ = R [1/2^2 - 1/n^2]
where λ is the wavelength of the spectral line, R is Rydberg constant and n is an integer greater than 2.
For the first line of Balmer series of hydrogen atom, n = 3. Therefore,
1/λ = R [1/2^2 - 1/3^2]
=> λ = 656.3 nm
The wavelength of the first line of Balmer series of hydrogen atom is 656.3 nm.
Now, let's consider the double ionized lithium atom (Li2+), with atomic number Z = 3.
As the lithium atom has lost two electrons, its effective nuclear charge is +3.
The formula for the wavelength of spectral lines of any atom is given by:
1/λ = Z^2 Rh [1/ni^2 - 1/nf^2]
where Rh is the Rydberg constant for the atom, ni and nf are the initial and final energy levels of the electron, and Z is the atomic number of the atom.
We need to find the wavelength of the first line of Balmer series for the Li2+ ion.
For the first line of Balmer series, nf = 2 and ni = 3.
Substituting the values in the above formula, we get:
1/λ = 3^2 Rh [1/3^2 - 1/2^2]
=> 1/λ = 9/36 Rh
=> λ = 4/9 Rh
The wavelength of the first line of Balmer series for the Li2+ ion is 4/9 times the wavelength of the first line of Balmer series for hydrogen.
Substituting the value of Rh for lithium (Rl) and Rh for hydrogen (R), we get:
λ = (4/9) (Rl/R) λ(H)
=> λ = (4/9) (Rl/R) (656.3 nm)
=> λ = (4/9) (4/9) λ(H)
=> λ = (16/81) λ(H)
=> λ = 656.3/9
=> λ = 72.92 nm
Therefore, the wavelength of the first line of Balmer series for the Li2+ ion is 72.92 nm, which is equal to λ(H)/9.
Hence, the correct answer is option (B) l/9.
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The wavelength of the first line of Balmer series of hydrogen atom is l Å. The wavelength of this line of a double ionised lithium atom Z = 3) isa)l/3 b) l/9c) l/8d) l/27Correct answer is option 'B'. Can you explain this answer?
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