calculate the wavelength of the second line in the Brackett series (nf=4) of the hydrogen emission spectrum. Rh= 2.180e-18
any help please? i know its energy levels 4 to 6... but that is all.
Not a presice
answer
To calculate the wavelength of the second line in the Brackett series (nf=4) of the hydrogen emission spectrum, we can use the Rydberg formula:
1/λ = R_H * (1/n_i^2 - 1/n_f^2)
Where:
λ is the wavelength of the emitted photon
R_H is the Rydberg constant (2.180e-18 m)
n_i is the initial state (6 in this case; corresponding to the bracket series)
n_f is the final state (4 in this case; corresponding to nf=4)
Substituting the values into the formula:
1/λ = 2.180e-18 * (1/6^2 - 1/4^2)
= 2.180e-18 * (1/36 - 1/16)
= 2.180e-18 * (16/576 - 36/576)
= 2.180e-18 * (-20/576)
= -3.795e-21
To calculate the wavelength, take the reciprocal:
λ = -1 / (-3.795e-21)
≈ 2.63e20 m
So, the wavelength of the second line in the Brackett series (nf=4) of the hydrogen emission spectrum is approximately 2.63 × 10^20 meters.
To calculate the wavelength of the second line in the Brackett series (nf=4) of the hydrogen emission spectrum, we can use the Rydberg formula. The Rydberg formula relates the energy levels of the hydrogen atom to the wavelength of the emitted light. It is given by:
1/λ = Rh * (1/nf^2 - 1/ni^2)
Where:
λ is the wavelength of the emitted light,
Rh is the Rydberg constant (2.180e-18),
nf is the final energy level (4, in this case),
and ni is the initial energy level.
In the Brackett series, the initial energy level ni ranges from 4 to infinity (or n = 4 to n = ∞). So we need to calculate the wavelength for each value of ni from 4 to 6 and find the wavelength of the second line (ni = 4).
Let's calculate the wavelength for each value of ni:
1/λ = Rh * (1/nf^2 - 1/ni^2)
For ni = 4:
1/λ = 2.180e-18 * (1/4^2 - 1/4^2)
1/λ = 2.180e-18 * (1/16 - 1/16)
1/λ = 0
As we can see, when ni = 4, the wavelength becomes infinite. This means that there is no second line in the Brackett series (nf=4) of the hydrogen emission spectrum.
1/lambda= R (1/4^2-1/662) R= Rydberg constant = (1.09678E7 m^-1)
A little algebra changes this to
1/lambda= R (36-16)/(16*36)
solve for lambda.