The energy levels the electron can occupy in the He+ ion can be calculated using the energy level equation. Calculate the frequency (Hz) of the photon emitted when the electron goes from a n= 4.00 energy level to a n= 3 energy level. Express answer in scientific notation.
1/wavelength = 2.180E-18*Z^2*(1/3^2-1/4^2)
Z for He is 2 so Z^2 = 4.
Find wavelength, then c = frequency x wavelength to convert to frequency.
do the (*) mean multiple..or are you using them to separate the numbers??
* means multiply.
To calculate the frequency of the photon emitted when the electron goes from a higher energy level to a lower energy level, you can use the Rydberg formula. The Rydberg formula represents the relationship between the energy levels of an electron in an atom and the frequency of the photon emitted:
1/λ = R * (1/n1^2 - 1/n2^2)
Where λ is the wavelength of the photon, R is the Rydberg constant (approximately 3.29 x 10^15 Hz), n1 is the initial energy level, and n2 is the final energy level.
Given that the initial energy level (n1) is 4.00 and the final energy level (n2) is 3, we can substitute these values into the formula and solve for the wavelength (λ), which can then be converted to frequency (Hz).
1/λ = R * (1/4.00^2 - 1/3^2)
1/λ = R * (1/16 - 1/9)
1/λ = R * (9/144 - 16/144)
1/λ = R * (-7/144)
λ = -144/7R
Now, we will substitute the value of R into the equation:
λ = -144/7 * (3.29 x 10^15 Hz)
λ ≈ -675.17 x 10^13 Hz
Since frequency is the inverse of wavelength, the frequency (f) can be calculated by taking the reciprocal of the wavelength:
f = 1/λ ≈ -1.481 x 10^-14 Hz
Therefore, the frequency of the photon emitted when the electron goes from n=4.00 to n=3 in the He+ ion is approximately -1.481 x 10^-14 Hz, expressed in scientific notation.