The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state

Q: Why does the atom recoil?

Just like a gun recoils when it fires a bullet, the atom must recoil when it 'fires' a photon to conserve linear momentum. Since the initial momentum of the atom was zero, the final momentum of the atom and the photon must add up to zero.

Q: Can we use the Rydberg formula instead of the energy formula?

Yes, you can use the Rydberg formula to find the wavelength $\lambda$, and then use $p = h/\lambda$. Both methods are physically equivalent and will yield the same result, but the energy method often involves fewer complex constants to multiply.

Q: What mass should be used for the hydrogen atom?

A hydrogen atom consists of one proton and one electron. The mass of the electron is negligible, so the mass of the hydrogen atom is approximately the mass of a proton, which is $1.67 \times 10^{-27}$ kg.

Q: Why did we convert electron volts to Joules?

To calculate the velocity in standard SI units (meters per second), all other quantities must also be in SI units. Electron volts (eV) is a unit of energy, but the SI unit is the Joule (J). The conversion factor is $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.

Q: Is the recoil energy of the atom significant enough to affect the photon's energy?

In highly precise calculations, the recoil energy of the atom does take a tiny fraction of the transition energy, slightly red-shifting the photon. However, because the atom is so much more massive than the photon's equivalent mass, this effect is negligible for standard textbook problems.

Question

The recoil speed of a hydrogen atom after it emits a photon in going from $n = 5$ state to $n =1$ state will be

Options:

(A) 4.17 m/s

(B) 2.19 m/s

(C) 3.25 m/s

(D) 4.34 m/s

Sigma Percentile · Accepted Answer

The Recoiling Atom: A Dance of Momentum and Light

Imagine a hydrogen atom floating perfectly still in the vast emptiness of space. Inside this tiny universe, an electron is sitting way up in the fifth energy level, the $n = 5$ state. It is excited, but like all things in nature, it seeks stability.

Now, this electron wants to return to its ground state, $n = 1$. When it makes this dramatic downward jump, it cannot just make the energy disappear. Instead, it releases its excess energy in the form of a brilliant flash—a photon.

### The Physics of the Recoil

This emitted photon shoots off at the speed of light. And even though a photon has absolutely no mass, it carries **momentum**. According to the principles of modern physics, the momentum of a photon is equal to its energy divided by the speed of light, $p = \frac{E}{c}$.

Here is where the physics gets truly beautiful. We must invoke the **Law of Conservation of Linear Momentum**. Since the hydrogen atom was initially at rest, the total initial momentum of our system was zero.

Therefore, the final momentum must also be zero. When the photon shoots off in one direction, the atom must recoil in the exact opposite direction to balance the scales. It is exactly like a gun recoiling when it fires a bullet! The momentum of the recoiling atom must perfectly equal the momentum of the emitted photon:
$$ p_{	ext{atom}} = p_{	ext{photon}} $$

### Calculating the Energy

To find out how fast the atom recoils, we first need to know the energy of the photon. We use the standard Bohr formula for the energy difference between two states:
$$ \Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV} $$

Substituting our energy levels, $n_1 = 1$ and $n_2 = 5$, we get:
$$ \Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{5^2} ight) 	ext{ eV} $$

Simplifying the fraction, we have $1 - \frac{1}{25}$, which is $\frac{24}{25}$. Multiplying this by $13.6$ gives us the exact energy released:
$$ \Delta E = 13.6 	imes \frac{24}{25} = 13.056 	ext{ eV} $$

### The Final Calculation

Now, we know that the recoil velocity of the atom is its momentum divided by its mass ($v = \frac{p}{m}$). And since the atom's momentum equals the photon's momentum ($\frac{E}{c}$), the velocity becomes:
$$ v = \frac{E}{mc} $$

Before we plug in the numbers, we must be careful! We need to convert the energy from electron volts to Joules to keep our units consistent. The conversion factor is $1 	ext{ eV} = 1.6 	imes 10^{-19} 	ext{ J}$.

The mass of a hydrogen atom is approximately the mass of a single proton, which is $1.67 	imes 10^{-27} 	ext{ kg}$. Let's substitute all these values into our velocity equation:
$$ v = \frac{13.056 	imes 1.6 	imes 10^{-19}}{1.67 	imes 10^{-27} 	imes 3 	imes 10^8} $$

Calculating this carefully, the powers of ten simplify beautifully, and we are left with a recoil speed of:
$$ v \approx 4.17 	ext{ m/s} $$

The atom drifts away at a gentle walking pace, a silent testament to the conservation laws that govern our universe.

Animated Solution for Physics - Atoms and Nuclei: The recoil speed of a hydrogen atom after it emits a photon in going from $n = 5$ state to $n = 1$ state will be

Select Answer:

Visualized Solution

$The Setup$

$Electron Transition$

$Photon Momentum$

$Conservation of Momentum$

$Energy of Emitted Photon$

$Calculating Energy$

$Recoil Velocity Formula$

$Final Substitution$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Physics of the Recoil

Calculating the Energy

The Final Calculation

Similar Questions