Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (

Q: Why does a photon have momentum if it has no mass?

Great question! While photons have zero *rest* mass, they carry energy and momentum by virtue of their wave nature. According to Einstein's relativity and quantum mechanics, a photon's momentum is directly related to its wavelength via the de Broglie equation $p = \frac{h}{\lambda}$.

Q: Why did we divide the molar mass by Avogadro's number?

The molar mass ($4\text{ g/mol}$) is the mass of one entire mole of Helium atoms, which is $6 \times 10^{23}$ atoms! Since only one single atom absorbs the photon, we need the mass of just that one atom. Dividing the mass of a mole by the number of atoms in a mole gives us the mass per atom.

Q: Why did we convert the mass from grams to kilograms?

Consistency is key in physics! Planck's constant is given in Joules seconds ($\text{J s}$). A Joule is defined as $\text{kg m}^2/\text{s}^2$. Because the Joule inherently uses kilograms, all other mass values in your equation must also be in kilograms to cancel out correctly.

Q: What happens to the energy of the photon after it's absorbed?

The photon's energy doesn't disappear. It is absorbed by the Helium atom, typically causing an electron to jump to a higher energy level (an excited state). A tiny fraction of that energy also goes into the kinetic energy of the recoiling atom, but the momentum transfer is what dictates the recoil velocity.

Q: Why did we multiply by 100 at the very end?

This is a classic trap! Our standard SI unit calculations gave us the velocity in meters per second ($\text{m/s}$). However, the question explicitly asked for the answer in centimeters per second ($\text{cm s}^{-1}$). Since $1\text{ m} = 100\text{ cm}$, we multiply by 100 to get the final correct numerical value.

Question

Consider a helium (He) atom that absorbs a photon of wavelength $330	ext{ nm}$. The change in the velocity (in $	ext{cm s}^{-1}$) of He atom after the photon absorption is ______.
(Assume: Momentum is conserved when photon is absorbed.
[Use: Planck constant $= 6.6 	imes 10^{-34}	ext{ J s}$, Avogadro number $= 6 	imes 10^{23}	ext{ mol}^{-1}$, Molar mass of He $= 4	ext{ g mol}^{-1}$]

Sigma Percentile · Accepted Answer

Have you ever wondered what happens when the quantum world collides with classical mechanics? Imagine a tiny, solitary Helium atom floating in the vast emptiness of space, perfectly at rest. Suddenly, a photon—a packet of pure light energy—comes hurtling towards it at the speed of light.

In a fraction of a nanosecond, the Helium atom absorbs the photon completely. But wait! The photon was moving, which means it carried momentum. According to the fundamental laws of the universe, momentum cannot simply vanish. It must be conserved. So, what happens to that momentum? It gets transferred entirely to the Helium atom, causing it to recoil and move.

This beautiful interplay between light and matter is exactly what we are going to unravel in this problem. We will bridge the gap between the wave nature of light and the kinetic motion of an atom. Let's break it down step by step.

### Analyzing the Setup: The Photon's Kick

Our journey begins with the incoming photon. Even though a photon has absolutely zero rest mass, it is not devoid of momentum. This is one of the most mind-bending revelations of quantum mechanics, brought to light by Louis de Broglie.

The de Broglie relation tells us that any wave has an associated momentum, and any particle has an associated wavelength. For our photon, this relationship is elegantly captured by the equation:

$$p_{	ext{photon}} = \frac{h}{\lambda}$$

Here, $h$ is Planck's constant, the fundamental pixel size of the quantum universe, and $\lambda$ is the wavelength of the light. The problem provides us with a wavelength of $330 	ext{ nm}$. Before we plug this into our equation, we must ensure our units are perfectly aligned. Physics is unforgiving when it comes to mismatched units! We convert nanometers to meters by multiplying by $10^{-9}$.

### Calculating the Photon's Momentum

Let's substitute the given values into our master equation. We have Planck's constant $h = 6.6 	imes 10^{-34} 	ext{ J s}$ and the wavelength $\lambda = 330 	imes 10^{-9} 	ext{ m}$.

$$p_{	ext{photon}} = \frac{6.6 	imes 10^{-34}}{330 	imes 10^{-9}}$$

To simplify this without getting lost in a sea of zeroes, let's look at the numbers first. $6.6$ divided by $330$ is exactly $0.02$. Now, let's handle the powers of ten. $10^{-34}$ divided by $10^{-9}$ is $10^{-25}$.

$$p_{	ext{photon}} = 0.02 	imes 10^{-25} 	ext{ kg m/s}$$

We can rewrite this in a much cleaner scientific notation by shifting the decimal point two places to the right, which decreases the exponent by two:

$$p_{	ext{photon}} = 2 	imes 10^{-27} 	ext{ kg m/s}$$

This incredibly tiny number represents the exact "kick" the photon delivers to the Helium atom upon absorption.

### Determining the Mass of a Single Helium Atom

Now that we know the force of the kick, we need to know how heavy the object being kicked is. The problem states that the molar mass of Helium is $4 	ext{ g/mol}$.

**Warning:** This is a classic trap! This is the mass of an entire *mole* of Helium atoms—that's $6 	imes 10^{23}$ atoms! We only care about one single, solitary atom.

To find the mass of one atom, we must divide the molar mass by Avogadro's number ($N_A$). Furthermore, to maintain harmony with our standard SI units (since Planck's constant uses kilograms), we must convert the $4 	ext{ grams}$ into $4 	imes 10^{-3} 	ext{ kilograms}$.

$$m_{	ext{He}} = \frac{	ext{Molar Mass}}{N_A}$$

$$m_{	ext{He}} = \frac{4 	imes 10^{-3} 	ext{ kg/mol}}{6 	imes 10^{23} 	ext{ mol}^{-1}}$$

Simplifying the fraction $4/6$ gives us $2/3$. Subtracting the exponents ($-3 - 23$) gives us $-26$.

$$m_{	ext{He}} = \frac{2}{3} 	imes 10^{-26} 	ext{ kg}$$

### The Master Equation: Conservation of Momentum

We have all our pieces on the chessboard. We know the momentum of the incoming photon, and we know the mass of the Helium atom. The grand principle of Conservation of Momentum dictates that the total momentum before the collision must equal the total momentum after the collision.

Since the Helium atom was initially at rest, the initial momentum is solely that of the photon. After absorption, the photon is gone, and the Helium atom is moving with a new velocity, $\Delta v$.

$$p_{	ext{initial}} = p_{	ext{final}}$$

$$p_{	ext{photon}} = p_{	ext{He}}$$

$$p_{	ext{photon}} = m_{	ext{He}} 	imes \Delta v$$

### Final Calculation: Bringing It All Together

Let's substitute our hard-earned values into this conservation equation:

$$2 	imes 10^{-27} = \left(\frac{2}{3} 	imes 10^{-26}ight) 	imes \Delta v$$

Now, we isolate $\Delta v$. We multiply both sides by $3$, divide by $2$, and divide by $10^{-26}$:

$$\Delta v = \frac{2 	imes 10^{-27} 	imes 3}{2 	imes 10^{-26}}$$

The $2$'s cancel out beautifully. We are left with $3 	imes 10^{-1}$, which is simply:

$$\Delta v = 0.3 	ext{ m/s}$$

**The Final Trap:** Are we done? Not quite! The examiners have laid one final snare. The question explicitly asks for the velocity in **$	ext{cm s}^{-1}$**.

To convert meters per second to centimeters per second, we multiply by $100$.

$$\Delta v = 0.3 	imes 100 = 30 	ext{ cm/s}$$

And there we have it! By carefully navigating quantum momentum, atomic mass, and unit conversions, we've successfully calculated the recoil velocity of the atom. The final answer is **30**.

Enter Numerical Value:

Visualized Solution

Visualizing the Collision

Momentum of a Photon

Calculating Photon Momentum

Evaluating Photon Momentum

Mass of a Single He Atom

Evaluating Mass

Conservation of Momentum

Substituting Values

Solving for Velocity

Final Answer in cm/s

The Way Forward

The Sigma Insight: Wave Particle Duality

Analyzing the Setup

Calculating the Photon's Momentum

Determining the Mass of a Single Helium Atom

The Master Equation

Final Calculation

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