Consider a sample of excited atoms that lie 3.211 × 10–19 J above the ground state. Determine the emission wavelength (in nanometers) of these atoms.

Well, well, well! Looks like we have some excited atoms here. They must be like my friend over there who's always jumping around. Now, to determine their emission wavelength, we can use a little equation called the "Rydberg formula."

The Rydberg formula is λ = hc / ΔE, where λ is the wavelength, h is Planck's constant (6.626 × 10^-34 J·s), c is the speed of light (3.00 × 10^8 m/s), and ΔE is the energy difference.

So, plugging in the values, we get:

λ = (6.626 × 10^-34 J·s * 3.00 × 10^8 m/s) / (3.211 × 10^-19 J)

Now let's do the math to find the wavelength. Bingo!

λ ≈ 619.6 nm

So, the emission wavelength of these excited atoms would be approximately 619.6 nanometers. Let's hope they don't emit any bad jokes along with it!

To determine the emission wavelength of the excited atoms, we can use the formula:

E = hc/λ

where:
E is the energy difference between the excited state and the ground state
h is Planck's constant (6.626 x 10^(-34) J·s)
c is the speed of light (3.00 x 10^8 m/s)
λ is the wavelength of the emitted light

First, we need to convert the given energy difference to joules:

Energy difference (E) = 3.211 x 10^(-19) J

Now, we can rearrange the formula to solve for the wavelength (λ):

λ = hc/E

Substituting the values:

λ = (6.626 x 10^(-34) J·s) * (3.00 x 10^8 m/s) / (3.211 x 10^(-19) J)

Calculating this, we get:

λ ≈ 6.156 x 10^(-7) m

To convert this to nanometers (nm), we multiply by 10^9:

λ (in nm) ≈ 6.156 x 10^(-7) m * (10^9 nm/m)

Simplifying this, we find:

λ (in nm) ≈ 615.6 nm

Therefore, the emission wavelength of these atoms is approximately 615.6 nanometers.

E = h f

3.2 * 10^-19 = 6.6*10^-34 f

f = .485 * 10^15 Hz

T = 1/f = 2.06 * 10-15 seconds

if c = 3*10^8 m/s

lambda =c T =3*10^8 * 2.06*10^-15

= 6.19 * 10^-7 meters

= 619 * 10^-9 meters

((6.626*10^-34)(3.00*10^8))/(3.332*10^-19)

= 5.966*10^-7
in nanometers it is 596.6 nm