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The first line in the Lyman series has wavelength λ. The wavelength of the first line in Balmer series is
- a)2/9 λ
- b)9/2 λ
- c)5/27 λ
- d)27/5 λ
Correct answer is option 'D'. Can you explain this answer?
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The first line in the Lyman series has wavelength λ. The wavelength of...
Explanation:
Lyman and Balmer series are the two important series of spectral lines in the hydrogen spectrum. The Lyman series lies in the ultraviolet region whereas the Balmer series lies in the visible region. The first line in the Lyman series has wavelength λ. We need to find the wavelength of the first line in the Balmer series.
The formula to calculate the wavelength of any spectral line in the hydrogen spectrum is given by:
$\frac{1}{\lambda} = R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
where R is the Rydberg constant and n1 and n2 are the principal quantum numbers such that n2 > n1.
The first line in the Lyman series corresponds to the transition from n2 = ∞ to n1 = 1. Substituting these values in the above formula, we get:
$\frac{1}{\lambda} = R(\frac{1}{1}-\frac{1}{\infty})$
$\frac{1}{\lambda} = R$
$\lambda = \frac{1}{R}$
Now, the first line in the Balmer series corresponds to the transition from n2 = ∞ to n1 = 2. Substituting these values in the above formula, we get:
$\frac{1}{\lambda'} = R(\frac{1}{2^2}-\frac{1}{\infty})$
$\frac{1}{\lambda'} = \frac{3}{4}R$
$\lambda' = \frac{4}{3R}$
Dividing λ' by λ, we get:
$\frac{\lambda'}{\lambda} = \frac{\frac{4}{3R}}{\frac{1}{R}}$
$\frac{\lambda'}{\lambda} = \frac{4}{3}$
Therefore, the wavelength of the first line in the Balmer series is 27/5 times the wavelength of the first line in the Lyman series. Hence, the correct option is D.
Lyman and Balmer series are the two important series of spectral lines in the hydrogen spectrum. The Lyman series lies in the ultraviolet region whereas the Balmer series lies in the visible region. The first line in the Lyman series has wavelength λ. We need to find the wavelength of the first line in the Balmer series.
The formula to calculate the wavelength of any spectral line in the hydrogen spectrum is given by:
$\frac{1}{\lambda} = R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
where R is the Rydberg constant and n1 and n2 are the principal quantum numbers such that n2 > n1.
The first line in the Lyman series corresponds to the transition from n2 = ∞ to n1 = 1. Substituting these values in the above formula, we get:
$\frac{1}{\lambda} = R(\frac{1}{1}-\frac{1}{\infty})$
$\frac{1}{\lambda} = R$
$\lambda = \frac{1}{R}$
Now, the first line in the Balmer series corresponds to the transition from n2 = ∞ to n1 = 2. Substituting these values in the above formula, we get:
$\frac{1}{\lambda'} = R(\frac{1}{2^2}-\frac{1}{\infty})$
$\frac{1}{\lambda'} = \frac{3}{4}R$
$\lambda' = \frac{4}{3R}$
Dividing λ' by λ, we get:
$\frac{\lambda'}{\lambda} = \frac{\frac{4}{3R}}{\frac{1}{R}}$
$\frac{\lambda'}{\lambda} = \frac{4}{3}$
Therefore, the wavelength of the first line in the Balmer series is 27/5 times the wavelength of the first line in the Lyman series. Hence, the correct option is D.
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The first line in the Lyman series has wavelength λ. The wavelength of the first line in Balmer series isa)2/9 λb)9/2 λc)5/27 λd)27/5 λCorrect answer is option 'D'. Can you explain this answer?
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