Find the energy (in joules) of the photon that is emitted when the electron in a hydrogen atom undergoes a transition from the n=7 energy level to produce a line in the Paschen series.
Paschen series are to n=3
n=7 so n^2 = 49
n=3 so n^2 = 9
use the Rydberg formula
1/lambda = Rh(1/9 - 1/49)
where Rh = 1.097 x 10^7 m^-1
then convert the wavelength (lambda) to energy using E=hf
To find the energy of a photon emitted during a transition in a hydrogen atom, we can use the formula:
E = (hc) / λ
where E is the energy of the photon, h is the Planck's constant (6.626 x 10^-34 J·s), c is the speed of light in vacuum (3.00 x 10^8 m/s), and λ is the wavelength of the emitted light.
To determine the wavelength of the emitted light in the Paschen series, we can use the formula:
1/λ = R * (1/n1^2 - 1/n2^2)
where R is the Rydberg constant (1.097 x 10^7 m^-1), and n1 and n2 are the initial and final energy levels, respectively.
Since the transition is from the n=7 energy level (n2) in the Paschen series, we can substitute n1=1 and n2=7 into the second equation to find λ. Plugging that value into the first equation will give us the energy of the photon.
1/λ = R * (1/1^2 - 1/7^2)
1/λ = R * (1 - 1/49)
1/λ = R * (48/49)
λ = 49/(48 * R)
Now, we can use this value of λ to find the energy (E) of the photon using the first equation:
E = (hc) / λ
E = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (49/(48 * R))
E = (6.626 x 3.00 x 10^-26) / (49/(48 * 1.097 x 10^7))
Calculating this expression will give us the energy (E) in joules.
To find the energy of the photon emitted during the transition from the n=7 energy level in a hydrogen atom to produce a line in the Paschen series, we can use the formula:
E = -R_H*(1/n_f^2 - 1/n_i^2)
Where E is the energy of the photon, R_H is the Rydberg constant for hydrogen (2.18 x 10^-18 J), and n_f and n_i are the final and initial energy levels, respectively.
For the Paschen series, the final energy level (n_f) is 3, and the initial energy level (n_i) is 7.
Plugging these values into the formula:
E = -2.18 x 10^-18 J * (1/3^2 - 1/7^2)
Simplifying the equation:
E = -2.18 x 10^-18 J * (1/9 - 1/49)
E = -2.18 x 10^-18 J * (40/441)
E = -8.05 x 10^-20 J
Since energy cannot be negative, we take the absolute value of the answer:
|E| = 8.05 x 10^-20 J
Therefore, the energy of the photon emitted during the transition is approximately 8.05 x 10^-20 joules.