An excited hydrogen atom emits light with a frequency of 1.141x10^14 Hz to reach the energy level for which n=4. In what principle quantum level did the electron begin?

Calculate wavelength from c = frequency*wavelength, then

1/w = 1.097E7(1/n - 1/4^2)
Solve for n. w must be in meters.

Well, let's see if we can shed some light on this question! When the hydrogen atom emits light, it's like a little disco party happening on a quantum level. Now, in this case, we know that the frequency of the emitted light is 1.141x10^14 Hz.

The energy levels in hydrogen are determined by the principle quantum number, n. So, we are given that the electron reached the energy level for which n = 4. Now, the frequency of the emitted light is directly related to the energy difference between the initial and final energy levels.

To calculate the initial energy level, we need to find the energy difference between n = 4 and the ground state, which is n = 1.

Now, for a moment, imagine the electron as a little astronaut jumping from one energy level to another. To go from n = 1 to n = 4, it has to jump up three energy levels. Each energy level in hydrogen has a specific energy value, given by the equation E = -13.6 eV/n^2, where eV stands for electron volt.

Let's calculate the energy difference between n = 1 and n = 4 by using this equation.

For n = 1, the energy is E1 = -13.6 eV/1^2 = -13.6 eV.
For n = 4, the energy is E4 = -13.6 eV/4^2 = -0.85 eV.

Now, we can calculate the energy difference between these two levels: ΔE = E4 - E1 = -0.85 eV - (-13.6 eV) = 12.75 eV.

Since the frequency of the emitted light is given as 1.141x10^14 Hz, we need to convert it to energy using the equation E = hf, where h is Planck's constant (6.626 x 10^-34 J s). Multiplying the frequency by Planck's constant gives us the energy of the emitted light.

Now, we can equate the energy difference ΔE to the energy of the emitted light (hf) and solve for n.

ΔE = hf
12.75 eV = hf

Solving for f:

f = ΔE/h = (12.75 eV) / (6.626 x 10^-34 J s) ≈ 1.929 x 10^15 Hz.

Uh-oh! It seems like there's a bit of a discrepancy here. The frequency we calculated (1.929 x 10^15 Hz) doesn't match the given frequency (1.141x10^14 Hz). So, either my calculations went a little haywire, or there's a mistake in the given numbers.

In conclusion, based on the given frequency of 1.141x10^14 Hz, it seems like there might be an error in the problem statement or given data. But hey, at least we had a fun little disco party while discussing quantum levels, right?

The energy levels of a hydrogen atom are given by the formula:

\(E = -\dfrac{{13.6 \, \text{{eV}}}}{{n^2}}\)

where \(E\) is the energy level in electron volts (eV) and \(n\) is the principal quantum number.

Given that the frequency of light emitted is \(1.141 \times 10^{14}\) Hz, we can use the relation:

\(E = h \cdot f\)

where \(E\) is the energy of a photon, \(h\) is Planck's constant (\(6.63 \times 10^{-34}\) J·s), and \(f\) is the frequency of light.

To convert the frequency to energy in Joules, we can rearrange the equation to solve for \(E\):

\(E = h \cdot f\)

\(E = (6.63 \times 10^{-34} \, \text{{J·s}}) \times (1.141 \times 10^{14} \, \text{{Hz}})\)

\(E = 7.56 \times 10^{-20} \, \text{{J}}\)

Next, we need to find the initial energy level. Since the final energy level is for \(n = 4\), we can set up an equation equal to the initial energy level (\(E_i\)) and solve for \(n_i\):

\(E_i = -\dfrac{{13.6 \, \text{{eV}}}}{{n_i^2}}\)

Converting the final energy to electron volts (eV):

\(E_f = 7.56 \times 10^{-20} \, \text{{J}}\)

\(E_f = (7.56 \times 10^{-20} \, \text{{J}}) \times \dfrac{{1 \, \text{{eV}}}}{{1.6 \times 10^{-19} \, \text{{J}}}}\)

\(E_f = 0.47 \, \text{{eV}}\)

Rearranging the equation for the initial energy level:

\(E_i = -\dfrac{{13.6 \, \text{{eV}}}}{{n_i^2}}\)

\(n_i^2 = -\dfrac{{13.6 \, \text{{eV}}}}{{E_i}}\)

\(n_i^2 = -\dfrac{{13.6 \, \text{{eV}}}}{{0.47 \, \text{{eV}}}}\)

\(n_i^2 = -29.0\)

Since the principal quantum number must be positive, we can ignore the negative sign:

\(n_i = \sqrt{29.0}\)

\(n_i \approx 5.39\)

Therefore, the electron began in the fifth principle quantum level.

To determine the principle quantum level (n1) from which the excited hydrogen atom started, we can use the formula for the frequency of light emitted during a transition from one energy level to another.

The formula relates the frequency (ν) of the emitted light to the initial and final energy levels:

ν = R_H * [(1/n1^2) - (1/n2^2)]

Where:
ν = frequency of the emitted light (1.141x10^14 Hz)
R_H = Rydberg constant for hydrogen (2.18x10^15 Hz)
n1 = initial energy level (unknown)
n2 = final energy level (4 in this case)

Rearranging the formula, we can solve for n1:

1/n1^2 = (ν/R_H) + (1/n2^2)

Let's plug in the given values:

1/n1^2 = (1.141x10^14 Hz) / (2.18x10^15 Hz) + (1/4^2)

Calculating the right side:

1/n1^2 ≈ 0.5237 + 0.0625

1/n1^2 ≈ 0.5862

To solve for n1, we'll take the reciprocal of both sides:

n1^2 = 1 / 0.5862

n1^2 ≈ 1.706

Taking the square root of both sides:

n1 ≈ √1.706

n1 ≈ 1.306

Rounding the result to the nearest whole number, we can conclude that the electron began in the principle quantum level (n1) approximately equal to 1.