Q.Monochromatic light of wavelength $632.8\ \text{nm}$ is produced by a helium-neon laser. The power emitted is $9.42\ \text{mW}$.

Concept: Photon Energy — each photon carries energy $E = h\nu = \frac{hc}{\lambda}$ and momentum $p = \frac{h}{\lambda}$.

(a)

Wavelength $\lambda = 632.8\ \text{nm} = 6.328 \times 10^{-7}\ \text{m}$.

Energy:

Momentum:

(b)

Power $P = 9.42\ \text{mW} = 9.42 \times 10^{-3}\ \text{J/s}$.

Number of photons per second:

(c) …

Concept and Intuition

A laser beam is a stream of photons. Even though the beam looks continuous, its power is the sum of the energies of individual photons arriving per second. Each photon, being a quantum of light, has energy proportional to its frequency and momentum inversely proportional to its wavelength — a direct consequence of de Broglie’s relation and Planck’s law.

The key is to treat the macroscopic power ($9.42\ \text{mW}$) as the product of the number of photons per second and the energy per photon. For part (c), we simply equate the photon’s momentum to the classical momentum of a hydrogen atom and solve for its speed.

Step-by-step solution

1. Photon energy from wavelength

The energy of a single photon is given by the Planck-Einstein relation:

where

$h = 6.626 \times 10^{-34}\ \text{J·s}$ (Planck’s constant),

$c = 3.00 \times 10^{8}\ \text{m/s}$,

$\lambda = 632.8\ \text{nm} = 632.8 \times 10^{-9}\ \text{m}$.

Substitute:

First compute numerator: $6.626 \times 3.00 = 19.878$, so $19.878 \times 10^{-26}\ \text{J·m}$.

Divide by $632.8 \times 10^{-9}$:

$\frac{19.878}{632.8} \approx 0.03141$, so

2. Photon momentum

For a photon, momentum is:

Substitute:

$\frac{6.626}{632.8} \approx 0.01047$, so

3. Number of photons per second

Power $P = 9.42\ \text{mW} = 9.42 \times 10^{-3}\ \text{J/s}$.

If each photon carries energy $E$, then the number of photons arriving per second is:

Substitute:

$n = \frac{9.42 \times 10^{-3}}{3.141 \times 10^{-19}}$ …

Method: Photon Energy–Momentum Relations and Power–Photon Rate Conversion

This problem uses two core ideas: the Planck–Einstein relations for a single photon, and the fact that the total power of a beam equals the energy per photon times the number of photons per second.

(a) Energy and momentum of each photon

Step 1 – Convert wavelength to metres.

$632.8\ \text{nm} = 632.8 \times 10^{-9}\ \text{m} = 6.328 \times 10^{-7}\ \text{m}$.

Step 2 – Energy of one photon.

Use $E = \frac{hc}{\lambda}$, where $h = 6.626 \times 10^{-34}\ \text{J·s}$ and $c = 3.00 \times 10^{8}\ \text{m/s}$.

First compute numerator: $6.626 \times 3.00 = 19.878$, with powers $10^{-34+8} = 10^{-26}$, so $1.9878 \times 10^{-25}\ \text{J·m}$.

Divide: $\frac{1.9878 \times 10^{-25}}{6.328 \times 10^{-7}} = 3.141 \times 10^{-19}\ \text{J}$.

Step 3 – Momentum of one photon.

Use $p = \frac{h}{\lambda}$.

(b) Number of photons per second arriving at the target

Step 1 – Power is energy per second.

$P = 9.42\ \text{mW} = 9.42 \times 10^{-3}\ \text{J/s}$.

Step 2 – Let $n$ be the number of photons per second.

Then $P = n \times E$ (energy of one photon).

Step 3 – Divide.

$9.42 / 3.141 \approx 3.00$, and $10^{-3} / 10^{-19} = 10^{16}$.

$n = 3.00 \times 10^{16}\ \text{photons/s}$ …

Here are the most common mistakes students make on this exact problem, and how to avoid each one.

1. Forgetting to convert units (nm → m, mW → W)

The most frequent error. Wavelength is given in nanometres ($1\ \text{nm} = 10^{-9}\ \text{m}$) and power in milliwatts ($1\ \text{mW} = 10^{-3}\ \text{W}$). Plugging $632.8$ directly into $E = hc/\lambda$ gives a completely wrong energy — off by nine orders of magnitude.

How to avoid: Before writing any formula, convert every quantity to SI base units. Write it explicitly:

Make this the first line of your solution.

2. Using the wrong formula for photon momentum

Many students write $p = mc$ or $p = E/c^2$ — both are incorrect for a photon. A photon has zero rest mass, so $p = mv$ does not apply.

How to avoid: Memorise the two equivalent forms for photon momentum:

Use the first one here since you already have $\lambda$. It’s direct and avoids any confusion with mass.

3. Confusing energy per photon with total power

Part (b) asks for the number of photons per second. A common mistake is to divide power by $c$ or by $\lambda$, or to use $E = hf$ but forget to find $f$ from $\lambda$.

How to avoid: Remember the chain:

So first compute $E_{\text{photon}} = hc/\lambda$, then:

Do not skip the intermediate step — write $E_{\text{photon}}$ explicitly before dividing.

4. Using the wrong mass for the hydrogen atom in part (c)

Some students use the mass of a proton ($1.67 \times 10^{-27}\ \text{kg}$) or the mass of an electron. The problem says hydrogen atom, so you need the mass of the entire atom — approximately the mass of a proton plus an electron, which is essentially $1.67 \times 10^{-27}\ \text{kg}$ (the electron contributes negligibly, but using the proton mass is acceptable here). A bigger mistake is using the mass of a hydrogen molecule ($H_2$) or forgetting mass entirely and trying $v = p/c$.

How to avoid: Write down the known mass of a hydrogen atom: …