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What is the ratio of maximum to minimum wavelength of lyman series?
The Lyman series corresponds to electronic transitions in hydrogen where the electron falls to the n=1 energy level. The maximum wavelength occurs when the transition is from n=2 (the first level above n=1), yielding a wavelength of approximately 121.6 nm. The minimum wavelength occurs when the transition is from n approaching infinity, resulting in a wavelength of 0.1 nm (or less). Therefore, the ratio of maximum to minimum wavelength for the Lyman series is about 1216:0.1 or 12160:1.
What color is the wavelength of light in the Balmer series that results from the transition of an electron from n 3 to n 2?
The wavelength of light in the Balmer series resulting from the transition of an electron from n=3 to n=2 corresponds to a color in the visible spectrum. Specifically, this transition emits light at a wavelength of approximately 656 nanometers, which falls within the red part of the spectrum. This transition is often referred to as the H-alpha line.
How do you find the relative maximum and minimum on your Ti Calculator?
If it is in the 83 series, try commands 3 and 4 on the CALC menu. Check your manual (or the online pdf manual) for more usage information.
How many complete picture frame are scan in one second?
30 frames per second is the minimum rate that it takes to fool the human eye into believing that a series of still pictures are moving.
What is the maximum product of n positive numbers if their sum is given?
Let the sum of series a1,.., an = A. Since ai >0. Then the maximum possible product of a1,..,an is = (A/n)n. This result basically comes the relation between the arithmetic mean and geometric mean of n positive numbers. A/n >= (a1...an)(1/n). The equality case of the above relation gives the maximum product (by raising the power by n on both sides).
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 ratio of maximum to minimum wavelength of lyman series?
The Lyman series corresponds to electronic transitions in hydrogen where the electron falls to the n=1 energy level. The maximum wavelength occurs when the transition is from n=2 (the first level above n=1), yielding a wavelength of approximately 121.6 nm. The minimum wavelength occurs when the transition is from n approaching infinity, resulting in a wavelength of 0.1 nm (or less). Therefore, the ratio of maximum to minimum wavelength for the Lyman series is about 1216:0.1 or 12160:1.
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.
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?
Well, the different series represent different electronic transitions. But there is an important equation, the Rydberg formula which describes all of them.. I think you've learned of it since you mention the n values. This lead to the Bohr model of the hydrogen atom, which explained _why_ you had these levels.Or, almost. See, it turned out that those lines were not actually single lines, but several lines very close together.. And so they had to add more variables to describe how these levels-within-levels fit together.. and the answer to that eventually came from quantum mechanics.
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.
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.
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?
Use the rydberg equation 1/wavelength = 109677[ 1/n one square - 1/n two square ] 109677 is in cm inverse for balmer series n one = 2 and for the fifth line n two = 7 putting them in the equation we get = 397 nm lies in the violet region of light
