# What does it mean by 'photons of higher intensity'?

I found an Assertion and Reason question related to transition energy in Bohr's orbits.

Assertion In the H-atom, photons of higher intensity are emitted when electron falls from the 2nd to 1st orbit than when it falls from the 4th to 2nd orbit.

Reason Transition energy from the 2nd to 1st orbit is greater than for the 4th to 2nd orbit.

The answer says the Assertion is false and Reason is true.

What does it mean by photons of higher intensity? Is it talking about the frequency? But if it was talking about the frequency, then the Assertion would be true.

• I don't think this is a well-formulated question. The Assertion is indeed false and the Reason is indeed true, but the Reason does not imply the falsehood of the Assertion; it is an unconnected observation. (Atomic orbital transitions can absorb or emit more than one photon, but only under special conditions.)
– zwol
Jul 17 '16 at 22:04

The assertion is incorrect. Intensity is related to the number of photons emitted and has nothing to do with Bohr orbit transitions.

The reason is correct. Rather than higher intensity, photons of higher energy are emitted when an electron falls from the 2nd to the 1st Bohr orbit as compared to when it falls from the 4th to the 2nd orbit. This is because the energy emitted or absorbed in a Bohr transition is given by the following equation where Z is the atomic number of the atom, R is a constant and n represents the quantum number of the final and initial Bohr orbits.

$\mathrm{\Delta E=-Z^2R \left ( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right )}$

since

$\mathrm{\frac{1}{1^2}-\frac{1}{2^2}=3/4}$

is greater than

$\mathrm{\frac{1}{2^2}-\frac{1}{4^2}=3/16}$

more energy is emitted in the first case than the second.

Ron's answer is completely correct, but I'm gonna give some more detail about what I think is the point of the assertion part:

The purpose of the assertion is to make you distinguish between the energy of classical waves and quantum mechanical/particle/electromagnetic waves. The assertion is wrong because the energy of a photon (or the energy of the wavy part of any particle) is proportional to the frequency of the photon as demonstrated in the photoelectric effect and for massive particles, using the de Broglie wavelength. On the other hand, classically, the energy in a mechanical wave, like a sound wave or an earthquake, is proportional to the square amplitude of the wave. The intensity of this classical wave is defined as, $$I_{classical}=\frac{Power}{Area}= \frac{Energy}{time*Area}\propto \frac{A^2}{time*Area}$$where $A$ is the amplitude of the wave, and we simply used the fact that power is defined to be energy per unit time. This basically means the intensity of a mechanical wave is the amount energy being transported by that wave to a specific area per unit time.

This way of thinking about intensity no longer makes sense though when you begin to look at electronic transitions. If the assertion were true, this would mean that a larger number of photons are being emitted in the $2\rightarrow1$ transitions than in the $4\rightarrow2$ transition. This would then mean that when we shoot photons at the hydrogen atom, we should see some larger number of photons being absorbed in $1\rightarrow2$ transition than in the $2\rightarrow4$ transition. This is not what we observe. Rather we see that only specific wavelengths (or frequencies) of light cause the transition to occur, and the number of photons being absorbed in the transition is one.

Hopefully that helps explain why they would put intensity in the assertion. Classically it makes some sense to think that would be correct, but this isn't a mechanical wave, and a distinction must be made there.

The answer is correct because intensity refers to the number of photons emitted (i.e. intensity of light) and not the frequency (inverse of wavelength) of the photon being emitted.

• Thank you so much for your answer... May I know how intensity of light is related to an atom (as per Bohr's model) Jul 17 '16 at 16:35
• If it spontaneously falls, 1 fall = 1 photon Jul 17 '16 at 16:37
• Oh I understand now. It is related with how the electron falls - straight to the said orbit or any permutation of the kind. But if that is the case, the light intensity of n=2 to n=1 maybe equal to n=4 to n=2 depending on if the electron falls straight to n=2. Is that right? Btw, you answered the original question, thanks :) Jul 17 '16 at 16:41
• Yes, your understanding is correct Jul 17 '16 at 16:54