when an electron in a hydrogen atom transitions from the n = 5 level to the n = 2 level, what wavelength photon (in nm) is emitted?
Thanks!
To find the wavelength of the photon emitted when an electron in a hydrogen atom transitions from the n = 5 level to the n = 2 level, we can use the Rydberg formula:
1/λ = R * (1/n₁² - 1/n₂²)
where λ is the wavelength of the photon, R is the Rydberg constant, and n₁ and n₂ are the initial and final energy levels, respectively.
Since we are given n₁ = 5 and n₂ = 2, we can substitute these values into the formula:
1/λ = R * (1/5² - 1/2²)
Simplifying this equation, we have:
1/λ = R * (1/25 - 1/4)
1/λ = R * (4 - 25)/100
1/λ = R * (-21)/100
1/λ = -21R/100
To find the value of the Rydberg constant, we can plug it into the equation:
R = 1.097 × 10⁷/m
Substituting this value into our equation, we get:
1/λ = -21(1.097 × 10⁷/m)/100
1/λ = -2.2997 × 10⁶/m
To calculate the wavelength, we can rearrange the equation:
λ = 1/(-2.2997 × 10⁶/m)
Calculating the value, we find:
λ ≈ 4.348 × 10⁻⁷ m
To convert this wavelength into nanometers (nm), we multiply by 10⁹:
λ ≈ 434.8 nm
Therefore, the wavelength of the photon emitted when the electron in a hydrogen atom transitions from the n = 5 level to the n = 2 level is approximately 434.8 nm.
To determine the wavelength of the photon emitted during the transition of an electron in a hydrogen atom, we can use the Rydberg formula. The Rydberg formula is given by:
\[ \frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]
Where:
- \( \lambda \) is the wavelength of the photon emitted (in meters)
- \( R \) is the Rydberg constant (\( 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( n_1 \) and \( n_2 \) are the initial and final energy levels of the electron, respectively.
In this case, the electron transitions from the \( n = 5 \) level to the \( n = 2 \) level. Plugging these values into the formula, we can solve for \( \lambda \):
\[ \frac{1}{\lambda} = R \left(\frac{1}{2^2} - \frac{1}{5^2}\right) \]
Simplifying further:
\[ \frac{1}{\lambda} = R \left(\frac{1}{4} - \frac{1}{25}\right) \]
\[ \frac{1}{\lambda} = R \left(\frac{21}{100}\right) \]
\[ \frac{1}{\lambda} = \frac{21R}{100} \]
\[ \lambda = \frac{100}{21R} \]
Substituting the value of \( R = 1.097 \times 10^7 \, \text{m}^{-1} \):
\[ \lambda = \frac{100}{21 \times 1.097 \times 10^7} \]
Evaluating this expression gives the wavelength \( \lambda \) in meters. To convert it to nanometers (nm), we need to multiply by \( 10^9 \):
\[ \lambda_{\text{nm}} = \lambda \times 10^9 \]
Calculating the value, we find:
\[ \lambda_{\text{nm}} \approx 656.46 \, \text{nm} \]
Therefore, when an electron transitions from the \( n = 5 \) level to the \( n = 2 \) level in a hydrogen atom, it emits a photon with a wavelength of approximately 656.46 nm.
1/wavelength = R*(1/2^2 - 1/5^2)
R = Rydberg constant = 1.097E7