If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series emission spectrum. Calculate the wavelength λ associated with the fifth line.
Express your answer with the appropriate units.
To calculate the wavelength associated with the fifth line of the Balmer series, we will use the Balmer formula:
1/λ = R_H * (1/n1² - 1/n2²)
where λ is the wavelength, R_H is the Rydberg constant for hydrogen (approximately 1.097 × 10^7 m⁻¹), n1 is the principal quantum number of the lower energy level, and n2 is the principal quantum number of the higher energy level.
For the Balmer series, the lower energy level (n1) is always 2 since the electron transitions occur to the second energy level. We're looking for the fifth line, which means that the electron is transitioning from the sixth energy level (n2 = 6) to the second energy level (n1 = 2).
1/λ = R_H * (1/2² - 1/6²)
1/λ = 1.097 × 10^7 * (1/4 - 1/36)
Now, we can calculate the numerical value:
1/λ = 1.097 × 10^7 * (0.25 - 0.0278)
1/λ = 1.097 × 10^7 * 0.2222
1/λ ≈ 2.436 × 10^6
Now, to find the wavelength, we can take the reciprocal of this value:
λ ≈ 4.106 × 10^-7 m
We can convert this to nanometers (nm) by multiplying by 10^9:
λ ≈ 410.6 nm
Therefore, the wavelength associated with the fifth line in the Balmer series emission spectrum is approximately 410.6 nm.
To calculate the wavelength of the fifth line in the Balmer series emission spectrum, we can use the Balmer series formula:
1/λ = R * (1/2^2 - 1/n^2)
where λ is the wavelength, R is the Rydberg constant, and n represents the energy level of the electron transition.
In this case, since we are looking for the fifth line, n is equal to 2 + 5 = 7.
Substituting the values into the formula:
1/λ = R * (1/2^2 - 1/7^2)
To simplify the equation:
1/λ = R * (1/4 - 1/49)
1/λ = R * (49/196 - 4/196)
1/λ = R * (45/196)
Now, we can solve for λ by taking the reciprocal of both sides:
λ = 196/45R
The value of the Rydberg constant, R, is approximately equal to 1.097 × 10^7 m^-1.
Substituting this value into the equation:
λ = 196/45 * (1.097 × 10^7 m^-1)
Calculating the value:
λ = 4.868 × 10^-7 m
Therefore, the wavelength associated with the fifth line in the Balmer series emission spectrum is approximately 4.868 × 10^-7 meters.
To calculate the wavelength associated with the fifth line in the Balmer series emission spectrum, we need to use the formula for the Balmer series:
1/λ = R_H * (1/4^2 - 1/n^2)
Where:
- λ represents the wavelength in meters
- R_H is the Rydberg constant for hydrogen, approximately equal to 1.097 * 10^7 m^-1
- n is the principal quantum number, which corresponds to the energy level of the electron in the hydrogen atom
In this case, we want to find the wavelength associated with the fifth line. This corresponds to n = 6 (since the nth line corresponds to n = n + 2). Plugging this value into the formula:
1/λ = (1.097 * 10^7 m^-1) * (1/4^2 - 1/6^2)
Simplifying further:
1/λ = (1.097 * 10^7 m^-1) * (1/16 - 1/36)
= (1.097 * 10^7 m^-1) * (20/360)
= (1.097 * 10^7 m^-1) * (1/18)
Now, invert both sides of the equation to solve for λ:
λ = 18/(1.097 * 10^7 m^-1)
≈ 1.641 \times 10^{-7} m
Therefore, the wavelength λ associated with the fifth line in the Balmer series emission spectrum is approximately 1.641 x 10^-7 meters.