Calculate the wavelength (in nanometers) of a photon emitted by a hydrogen atom when its electron drops from the n = 4 state to the n = 3 state.
1/wavelength = R*(1/N1^2 - 1/N2^2)
R is the Rydberg constant and is 1.0973732E7/meter.
Wavelength has the unit of meters. You must change to nanometers. N1 = 3 and N2 = 4
Calculate the wavelength of the photon emitted when an electron makes a transition from to . You can make use of the following constants:
To calculate the wavelength of a photon emitted when an electron drops from one energy level to another in a hydrogen atom, we can use the Rydberg formula:
1/λ = R * (1/n₁² - 1/n₂²)
Where:
λ is the wavelength of the emitted photon,
R is the Rydberg constant (approximately 1.097 × 10^7 m⁻¹),
n₁ is the initial energy level,
n₂ is the final energy level.
Given that the electron drops from the n = 4 state to the n = 3 state, we substitute n₁ = 4 and n₂ = 3 into the formula.
To calculate the wavelength of a photon emitted by a hydrogen atom, we can use the formula:
\[ \frac{1}{\lambda} = R_H \left(\frac{1}{{n_1}^2} - \frac{1}{{n_2}^2}\right) \]
Where:
- λ (lambda) is the wavelength of the emitted photon,
- RH is the Rydberg constant (approximated as 1.097 x 10^7 m⁻¹),
- n1 is the initial energy level of the electron,
- n2 is the final energy level of the electron.
In this case, the electron drops from n = 4 to n = 3, so we can substitute these values into the formula:
\[ \frac{1}{\lambda} = R_H \left(\frac{1}{4^2} - \frac{1}{3^2}\right) \]
Now we can solve for the wavelength:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{16} - \frac{1}{9}\right) \]
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{9 - 16}{144}\right) \]
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{-7}{144}\right) \]
\[ \frac{1}{\lambda} = -6.474 \times 10^4 \]
To find the wavelength, we take the reciprocal:
\[ \lambda = \frac{1}{-6.474 \times 10^4} \]
\[ \lambda \approx - 1.545 \times 10^{-5} \]
It is important to note that this value is negative, which is due to the convention of the formula. However, since wavelength cannot be negative, we take the absolute value:
\[ \lambda \approx 1.545 \times 10^{-5} \]
Finally, since the question asks for the wavelength in nanometers (nm), we convert from meters to nanometers by multiplying by \(10^{9}\):
\[ \lambda \approx 1.545 \times 10^{-5} \times 10^9 \approx 154.5 \, \text{nm} \]
Therefore, the wavelength of the photon emitted by the hydrogen atom is approximately 154.5 nm.