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.
Z for He is 2 so Z^2 = 4.
Find wavelength, then c = frequency x wavelength to convert to frequency.
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.