An electron in the n=7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 2166 nm.
What is the value of n for the level to which the electron relaxed?
To determine the value of n for the level to which the electron relaxed, we can use the Rydberg formula:
1/λ = R( (1/n₁²) - (1/n₂²) )
where λ is the wavelength of the emitted light, R is the Rydberg constant, and n₁ and n₂ are the initial and final energy levels, respectively.
Given that the emitted light has a wavelength of 2166 nm, we can convert it to meters by dividing by 10^9:
λ = 2166 nm = 2166 x 10^-9 m
The Rydberg constant is approximately 1.097 x 10^7 m⁻¹.
Using the equation and plugging in the values:
1/λ = R( (1/n₁²) - (1/n₂²) )
1/(2166 x 10^-9 m) = (1.097 x 10^7 m⁻¹)( (1/7²) - (1/n₂²) )
Simplifying the equation:
2166 x 10^9 = (1.097 x 10^7)( (1/49) - (1/n₂²) )
Now, let's solve for n₂ by rearranging the equation:
(1/49) - (1/n₂²) = (2166 x 10^9) / (1.097 x 10^7)
(writing the quotient in scientific notation):
(1/49) - (1/n₂²) = 1.978 x 10^2
To simplify further, let's find a common denominator:
[(1)(n₂²) - (49)(1)] / (49)(n₂²) = 1.978 x 10^2
n₂² - 49 = (1.978 x 10^2)(49)(n₂²)
n₂² - 49 = 96.822 x 10^2 (n₂²)
Rearranging the equation:
96.822 x 10^2 (n₂²) - n₂² = 49
Combining like terms:
96.822 x 10^2 (n₂²) - 1(n₂²) = 49
Multiplying through by 100 to remove the decimal:
9682.2(n₂²) - 1(n₂²) = 4900
Combining like terms:
9681.2(n₂²) = 4900
Dividing both sides by 9681.2:
(n₂²) = 4900 / 9681.2
(n₂²) = 0.5066
Taking the square root of both sides:
n₂ = √0.5066
n₂ ≈ 0.71
Since n must be a positive integer, the value of n for the level to which the electron relaxed is approximately n = 1.
To determine the value of n for the level to which the electron relaxed, we can use the equation for the energy of a hydrogen atom:
E = -13.6 eV / n^2
Where E is the energy of the level and n is the principal quantum number.
We can first find the initial energy of the electron in the n=7 level by substituting n = 7 into the equation:
E_initial = -13.6 eV / (7^2)
= -13.6 eV / 49
Next, we can use the energy of the emitted light to find the final energy of the electron. The energy of a photon is given by:
E_photon = hc / λ
Where h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength of light (2166 nm = 2.166 x 10^-6 m).
E_final = hc / λ
= (6.626 x 10^-34 J s)(3.0 x 10^8 m/s) / (2.166 x 10^-6 m)
Now, we can equate the initial and final energies to find the value of n for the level to which the electron relaxed:
-13.6 eV / 49 = (6.626 x 10^-34 J s)(3.0 x 10^8 m/s) / (2.166 x 10^-6 m)
To solve for n, we can rearrange the equation and solve for n^2:
n^2 = (49)(-13.6 eV) / [(6.626 x 10^-34 J s)(3.0 x 10^8 m/s) / (2.166 x 10^-6 m)]
Finally, we can take the square root of both sides of the equation to find the value of n:
n = √[(49)(-13.6 eV) / [(6.626 x 10^-34 J s)(3.0 x 10^8 m/s) / (2.166 x 10^-6 m)]]
Evaluating this expression will give us the value of n for the level to which the electron relaxed.
1/wavelength = R(1/x^2 - 1/n^2)
R is the Rydberg constant. I assume you have that in your tables. Remember wavelength must be in meters. n is 7