Question

The visible emission lines observed by Balmer all involved nf = 2.

Which of the following is the best explanation of why the lines with nf = 3 are not observed in the visible portion of the spectrum?

Transitions to nf = 3 emit photons in the ultraviolet portion of the spectrum.

Transitions to nf = 3 emit photons in the infrared portion of the spectrum.

Transitions to nf = 3 emit photons that are at exactly the same wavelengths as those to nf = 2.

Transitions to nf = 3 are not allowed to happen.

Calculate the wavelength of the line in the Balmer series for which ni = 3.

Calculate the wavelength of the line in the Balmer series for which ni = 4.

Calculate the wavelength of the line in the Balmer series for which ni = 5.

Answer 1

1)

for nf = 3, energy is less than nf = 2

So, wavelength of photons for nf=3 will be more than that visible

This is infrared radiation

Transitions to nf = 3 emit photons in the infrared portion of the spectrum.

2)

Here photon will be emitted

1/lambda = R* (1/nf^2 - 1/ni^2)

R is Rydberg constant. R = 1.097*10^7

1/lambda = R* (1/nf^2 - 1/ni^2)

1/lambda = 1.097*10^7* (1/2^2 - 1/3^2)

lambda = 6.563*10^-7 m

lambda = 656 nm

Answer: 656 nm

3)

Here photon will be emitted

1/lambda = R* (1/nf^2 - 1/ni^2)

R is Rydberg constant. R = 1.097*10^7

1/lambda = R* (1/nf^2 - 1/ni^2)

1/lambda = 1.097*10^7* (1/2^2 - 1/4^2)

lambda = 4.862*10^-7 m

lambda = 486 nm

Answer: 486 nm

4)

Here photon will be emitted

1/lambda = R* (1/nf^2 - 1/ni^2)

R is Rydberg constant. R = 1.097*10^7

1/lambda = R* (1/nf^2 - 1/ni^2)

1/lambda = 1.097*10^7* (1/2^2 - 1/5^2)

lambda = 4.341*10^-7 m

lambda = 434 nm

Answer: 434 nm