This can be calculated using the following procedure: 1 / w = (Rh / hc)(1/nf^2 - 1/ni^2); 1 / w = ((2.18 x 10^-18) / (6.63 x 10^-14)(3.00 x 10^8))(1/(1^2) - 1/(6^2)); 1 / w = (1.10 x 10^7)(0.972); 1 / w = 1.07 x 10^7; w = 9.35 x 10^-8 m = 93.5 nm. The emission by the hydrogen atom is of wavelength 93.5 nm.
2.48 x 10^15 Hz. The energy of these photons is 1.64 x 10^-18 J (-5.4 x 10^-19 J - -2.18x 10^-18 J). Using equation 8.5, the frequency is found to be 2.48 x 10^15 Hz by dividing the photon energy by Planck's Constant (1.64 x 10^-18 J / 6.63 x 10^-34 J*s). They are ultraviolet.
A bit dicy for me since it has been a long time, but I have the formula sheet. Attend.
This formula to find Hertz: ( v = Hertz )
v = (3.29 X 10^15 s^-1) * Z^2 * (1/n final - 1/n initial )
Z = 1 for hydrogen
v = ( 3.29 X 10^15 ) * 1^2 * (1/2^2 - 1/4^2 )
= 6.169 X 10^14 Hertz
Now to convert Hertz into wavelength use this; Wavelength = Speed of light/Hertz
2.998 X 10^8/6.169 X 10^14 = 4.86 X 10^-7 meters (speed of light has m/s units while Hertz has s units, so meters come out )
to get nanometers; 4.86 X 10^-7 times 10^9 = 486 nanometers of wavelength.
( look up that color of visable light )
Been awhile, so long answer.
a) Frequency (v) = (3.29 X 10^15 s^-1)(Z^2)(1/Nf^2-1/Ni^2)
Z = 1 for H
Hertz = (3.29 X 10^15)(1^2)(1/2^2-1/3^2)
= 4.569 X 10^14 Hertz
Energy = Planck's constant * frequency in hertz
= (6.626 X 10^-34 J*s)(4.569 X 10^14 Hertz)
= 3.027 X 10^-19 Joules
b) Energy = Planck's constant * speed of light/wavelength in meters
3.027 X 10^-19 J = (6.626 X 10^-34)(2.998 X 10^8 m/s)/wavelength
= 6.563 X 10^-7 meters
which is 656.3 nanometers
You need to look up the Rydberg equation, which can be used to solve problems of this form.
The typesetting functionality here isn't really good enough to show the equation, so I've added a link in the Related Links section to the Wikipedia article on it.
The end value of "n" is 2.
A transition from 4p to 3p will produce light with a longer wavelength. This is because this transition is a smaller energy exchange than that of 3p to 2s (longer wavelength = less energy.)
Lyman
dd
With reference to the wikipedia article on this topic: The Balmer series predicts visible light wavelengths with high accuracy. The limiting transition wavelength predicted by the formula, inf -> 2, would be 364.6 nm.
The n4-n2 transition of hydrogen is in the cyan, with wavelength of 486.1 nm. blue = als
The end value of "n" is 2.
beta, aka an electron.
A transition from 4p to 3p will produce light with a longer wavelength. This is because this transition is a smaller energy exchange than that of 3p to 2s (longer wavelength = less energy.)
Yes. Although hydrogen is a non-metal, there is metallic hydrogen. It is formed when hydrogen is sufficiently compressed and undergoes a phase change; it is an example of degenerate matter. Solid metallic hydrogen consists of a crystal lattice of protons with a spacing which is significantly smaller than a Bohr radius. Indeed, the spacing is more comparable with an electron wavelength. The electrons are unbound and behave like the conduction electrons in a metal. As is the dihydrogen molecule H2, metallic hydrogen is an allotrope. In liquid metallic hydrogen, protons do not have lattice ordering.
That is the "gamma" line of the Lyman series: 94.97 nanometers.
transition of an electron from a higher energy level to a lower energy level.
Lyman
No, hydrogen is a gas.
dd
Hydrogen electron configuration will be 1s1.
Hydrogen is an element, the electron is a subatomic particle.