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What Is the ratio of minimum and maximum wavelength in balmer series?
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What is ratio of maximum and minimum wavelength of balmer series?
5:9 ,i am not sure (;
What is the maximum wavelength of balmer series?
The Balmer series is a series of spectral lines in the hydrogen spectrum that corresponds to transitions from energy levels n > 2 to the n=2 level. The longest wavelength in the Balmer series corresponds to the transition from n = ∞ to n = 2, known as the Balmer limit, which is approximately 656.3 nm.
What is the shortest wavelength radiation in balmer series?
The shortest wavelength radiation in the Balmer series is the transition from the n=3 energy level to the n=2 energy level, which corresponds to the Balmer alpha line at 656.3 nm in the visible spectrum of hydrogen.
What is the wavelength of the hydrogen atom in the 2nd line of the Balmer series?
The wavelength of the hydrogen atom in the 2nd line of the Balmer series is approximately 486 nm. This corresponds to the transition of an electron from the third energy level to the second energy level in the hydrogen atom.
What color is the wavelength of light in the balmer series that results from the transition of an electron from n4 to n2?
The n4-n2 transition of hydrogen is in the cyan, with wavelength of 486.1 nm. blue = als
What color is the wavelength of light in the Balmer series that results from the transition of an electron to N equals 2?
There are 4 Balmer lines with wavelengths in the visible region. They are red, aqua and two shades of violet. Other Balmer lines are in the ultraviolet. The red line corresponds to the transition from n = 3 to n = 2, the subsequent ones are from the 4, 5 and 6 levels to n = 2.
What are the colours in the balmer series?
The Balmer series consists of visible spectral lines emitted by hydrogen atoms when electrons transition from higher to lower energy levels. The colors in the Balmer series include red (656.3 nm), blue-green (486.1 nm), and violet (434.0 nm) wavelengths.
Spectral lines of the Lyman and Balmer series do not overlap Verify this statement by calculating the longest wavelength associated with the Lyman series and shortest wavelength associated with the B?
The longest wavelength in the Lyman series is the transition to n=2, which corresponds to the Lyman-alpha line at 121.6 nm. The shortest wavelength in the Balmer series is the transition to n=2, which corresponds to the Balmer-α line at 656.3 nm. Since the Lyman-alpha line has a longer wavelength than the Balmer-α line, they do not overlap.
If your eyes could see a slightly wider region of the electromagnetic spectrum you would see a fifth line in the Balmer series emission spectrum Calculate the wavelength lambda associated with the fif?
The Balmer series for hydrogen consists of four spectral lines in the visible region. If there were a fifth line, its wavelength could be calculated using the formula 1/λ = RH(1/4^2 - 1/n^2), where RH is the Rydberg constant and n is the energy level. Plugging in the values, the fifth line wavelength would be smaller than the existing lines in the series.
What is the name of the visible series in the hydrogen spectrum?
I believe it to be the Balmer Series.
If your eyes could see a slightly wider region of the electromagntic spectrum we would see a fifth line in the balmer series emissions spectrum what is the wavelenght associted with the fifth line?
The Balmer series corresponds to the spectral lines produced by hydrogen when electrons transition to the second energy level. The formula to calculate the wavelength for each line in the Balmer series is 1/λ = R(1/2^2 - 1/n^2), where n is the energy level and R is the Rydberg constant. For the fifth line (n=6), the wavelength would be approximately 434 nm.
In hydrogen spectrumwhat is the ratio of first line of Lyman series to the first line of balmer series?
The ratio of the first line of the Lyman series to the first line of the Balmer series in the hydrogen spectrum is 1:5.
