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The wavelength of the first line of Lyman series for hydrogen atom is equal to that
of the second line of Balmer series for a hydrogen like ion. The atomic number Z of
hydrogen like ion is
of the second line of Balmer series for a hydrogen like ion. The atomic number Z of
hydrogen like ion is
- a)2
- b)3
- c)4
- d)1
Correct answer is option 'A'. Can you explain this answer?
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Explanation:
The Lyman series involves transitions of electrons to the first energy level (n=1), while the Balmer series involves transitions to the second energy level (n=2). The formula for calculating the wavelength of a spectral line is given by:
1/λ=RZ^2(1/n1^2-1/n2^2)
where R is the Rydberg constant, Z is the atomic number, n1 and n2 are the initial and final energy levels of the electron.
Given that the wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion, we can write:
1/λLyman = R(1/n1^2 - 1/1^2)
1/λBalmer = RZ^2(1/n1^2 - 1/2^2)
Since the wavelengths are equal, we can equate the two expressions above:
R(1/n1^2 - 1/1^2) = RZ^2(1/n1^2 - 1/2^2)
Simplifying this equation, we get:
1/n1^2 = (4Z^2 - 1)/4
Since n1 is an integer, the only possible value for Z is 2. Therefore, the atomic number Z of the hydrogen like ion is 2, which corresponds to helium.
Answer: (A) 2
The Lyman series involves transitions of electrons to the first energy level (n=1), while the Balmer series involves transitions to the second energy level (n=2). The formula for calculating the wavelength of a spectral line is given by:
1/λ=RZ^2(1/n1^2-1/n2^2)
where R is the Rydberg constant, Z is the atomic number, n1 and n2 are the initial and final energy levels of the electron.
Given that the wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion, we can write:
1/λLyman = R(1/n1^2 - 1/1^2)
1/λBalmer = RZ^2(1/n1^2 - 1/2^2)
Since the wavelengths are equal, we can equate the two expressions above:
R(1/n1^2 - 1/1^2) = RZ^2(1/n1^2 - 1/2^2)
Simplifying this equation, we get:
1/n1^2 = (4Z^2 - 1)/4
Since n1 is an integer, the only possible value for Z is 2. Therefore, the atomic number Z of the hydrogen like ion is 2, which corresponds to helium.
Answer: (A) 2
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The wavelength of the first line of Lyman series for hydrogen atom is equal to thatof the second line of Balmer series for a hydrogen like ion. The atomic number Z ofhydrogen like ion isa)2b)3c)4d)1Correct answer is option 'A'. Can you explain this answer?
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