The visible region of the hydrogen spectrum was first studied by
A. Lyman
B. Balmer
C. Pfund
D. Brakett
Answer
Verified
471.3k+ views
Hint: The hydrogen spectrum consists of certain wavelengths, which lie in the visible region of the electromagnetic spectrum. These rays were observed by a scientist who gave a formula for finding the value of wavelength for these visible radiations and the series were named after him.
Complete step-by-step solution:
When an electron in the hydrogen atom is excited to higher energy levels then it can de-excite and lose energy jumping to lower energy levels. The energy is emitted in the form of electromagnetic radiation of different wavelengths. These wavelengths lie over the infrared, visible, and ultraviolet region of the electromagnetic spectrum. The wavelengths in the range from 400 nm to 800 nm lie in the visible region and are known as the Balmer series in the hydrogen spectrum since these spectral lines were first discovered and studied by Balmer. He discovered the pattern in these visible radiations emitted by hydrogen and had given a formula based on his studies for finding the wavelength of light emitted by the hydrogen atom.
Based on this discussion, the correct answer is option B.
Additional information:
The formula for wavelength is given as
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{n^2}}}} \right)$
Here $\lambda $ represents the wavelength of light emitted by the hydrogen atom in the visible region. R is known as the Rydberg’s constant whose value is given as
$R = 109677{m^{ - 1}}$
The parameter n signifies the level above the first excited state from which the electron de-excites to the first excited state corresponding to n = 2. The value of n is always greater than 2.
Note: The Balmer series is only one part of the spectrum emitted by a hydrogen atom. There are radiations emitted, which lie in the infrared and ultraviolet region as well. The Balmer’s formula can be generalized to the following expression which can give wavelength for all de-excitations in the hydrogen atom.
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Here ${n_1}$ is the lower energy level while ${n_2}$ is the higher energy level.
Complete step-by-step solution:
When an electron in the hydrogen atom is excited to higher energy levels then it can de-excite and lose energy jumping to lower energy levels. The energy is emitted in the form of electromagnetic radiation of different wavelengths. These wavelengths lie over the infrared, visible, and ultraviolet region of the electromagnetic spectrum. The wavelengths in the range from 400 nm to 800 nm lie in the visible region and are known as the Balmer series in the hydrogen spectrum since these spectral lines were first discovered and studied by Balmer. He discovered the pattern in these visible radiations emitted by hydrogen and had given a formula based on his studies for finding the wavelength of light emitted by the hydrogen atom.
Based on this discussion, the correct answer is option B.
Additional information:
The formula for wavelength is given as
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{n^2}}}} \right)$
Here $\lambda $ represents the wavelength of light emitted by the hydrogen atom in the visible region. R is known as the Rydberg’s constant whose value is given as
$R = 109677{m^{ - 1}}$
The parameter n signifies the level above the first excited state from which the electron de-excites to the first excited state corresponding to n = 2. The value of n is always greater than 2.
Note: The Balmer series is only one part of the spectrum emitted by a hydrogen atom. There are radiations emitted, which lie in the infrared and ultraviolet region as well. The Balmer’s formula can be generalized to the following expression which can give wavelength for all de-excitations in the hydrogen atom.
$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Here ${n_1}$ is the lower energy level while ${n_2}$ is the higher energy level.
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