Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The

Q: Why do we use the mass of a proton for the hydrogen atom?

A hydrogen atom consists of one proton and one electron. Since the mass of an electron is about 1836 times smaller than that of a proton, the total mass of the atom is overwhelmingly dominated by the proton. Thus, using the proton's mass is a very accurate approximation.

Q: Why is the change in angular momentum negative?

The negative sign indicates a decrease in the system's angular momentum. As the electron transitions from a higher orbit ($n=4$) to a lower orbit ($n=2$), it moves closer to the nucleus and its orbital angular momentum decreases.

Q: How do we know the photon momentum is $E/c$?

According to the theory of relativity, the energy and momentum of a massless particle like a photon are related by the equation $E = pc$. Rearranging this gives $p = E/c$.

Q: Why do we convert eV to Joules in the final step?

To calculate the recoil speed in standard SI units (meters per second), all other quantities in the equation $v = E/mc$ must also be in SI units. Therefore, the energy must be in Joules, mass in kilograms, and speed of light in meters per second.

Q: Can the electron transition to $n=1$ instead?

No, because the energy difference between any state and the ground state ($n=1$) is at least $10.2 \text{ eV}$ (from $n=2$ to $n=1$). Since our photon only has $2.55 \text{ eV}$ of energy, it cannot possibly involve a transition to the ground state.

Question

Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV.
Find
(a) the energy of the photons causing the photoelectrons emission.
(b) the quantum numbers of the two levels involved in the emission of these photons.
(c) the change in the angular momentum of the electron in the hydrogen atom, in the above transition, and
(d) the recoil speed of the emitting atom assuming it to be at rest before the transition. (Ionization potential of hydrogen is 13.6 eV.)

Sigma Percentile · Accepted Answer

### The Photoelectric Connection

Imagine a beam of light traveling from a hydrogen discharge tube and striking a piece of sodium metal. This light carries energy, and when it hits the sodium surface, it knocks out electrons. This is the classic **Photoelectric Effect**.

We are given that the fastest of these ejected electrons has a kinetic energy of $0.73 	ext{ eV}$, and the work function of sodium—the minimum energy required to just pull an electron out—is $1.82 	ext{ eV}$.

According to Einstein's photoelectric equation, the energy of the incident photon is simply the sum of the work function and the maximum kinetic energy of the emitted electron.

$$E = K_{\max} + \Phi$$

Substituting our values, we get:

$$E = 0.73 	ext{ eV} + 1.82 	ext{ eV} = 2.55 	ext{ eV}$$

This tells us that the photons arriving from the hydrogen tube carry exactly $2.55 	ext{ eV}$ of energy.

### Decoding the Hydrogen Spectrum

Now, we must ask: how did the hydrogen atom produce a photon of exactly $2.55 	ext{ eV}$?

In the Bohr model of the hydrogen atom, electrons exist in discrete energy levels given by the formula:

$$E_n = -\frac{13.6}{n^2} 	ext{ eV}$$

Let's calculate the energy of the first few levels. For the ground state ($n=1$), it's $-13.6 	ext{ eV}$. For the first excited state ($n=2$), it's $-3.4 	ext{ eV}$. For $n=3$, it's $-1.51 	ext{ eV}$, and for $n=4$, it's $-0.85 	ext{ eV}$.

When an electron jumps from a higher energy level to a lower one, it emits a photon whose energy equals the difference between those levels. We need to find two levels whose difference is exactly $2.55 	ext{ eV}$.

Let's test the transition from $n=4$ to $n=2$:

$$\Delta E = E_4 - E_2 = -0.85 	ext{ eV} - (-3.4 	ext{ eV}) = 2.55 	ext{ eV}$$

Perfect! The photon was emitted due to an electron transitioning from the $n=4$ state to the $n=2$ state.

### The Angular Momentum Shift

Bohr's second postulate states that the angular momentum of an electron in the $n^{	ext{th}}$ orbit is quantized and given by:

$$L = \frac{nh}{2\pi}$$

We need to find the change in angular momentum during this transition. The change is the final angular momentum minus the initial angular momentum.

$$\Delta L = L_f - L_i = L_2 - L_4$$

Substituting the values of $n$:

$$\Delta L = \frac{2h}{2\pi} - \frac{4h}{2\pi} = -\frac{2h}{2\pi} = -\frac{h}{\pi}$$

The negative sign beautifully illustrates that the electron has lost angular momentum as it dropped to a lower, more tightly bound orbit.

### The Recoil of the Atom

Finally, let's consider the physical mechanics of the emission. When the hydrogen atom emits a photon, it acts like a tiny cannon firing a cannonball. To conserve linear momentum, the atom must recoil in the opposite direction.

The momentum of the emitted photon is given by its energy divided by the speed of light:

$$p_{	ext{photon}} = \frac{E}{c}$$

By the **Conservation of Linear Momentum**, the recoil momentum of the atom must equal the momentum of the photon:

$$m v = \frac{E}{c}$$

We can rearrange this to solve for the recoil speed $v$:

$$v = \frac{E}{mc}$$

Now, we must be careful with units. We need to convert the photon energy from electron volts to Joules by multiplying by $1.6 	imes 10^{-19}$. The mass of a hydrogen atom is approximately the mass of a single proton, which is $1.67 	imes 10^{-27} 	ext{ kg}$.

$$v = \frac{2.55 	imes 1.6 	imes 10^{-19}}{1.67 	imes 10^{-27} 	imes 3 	imes 10^8}$$

Calculating this yields:

$$v \approx 0.814 	ext{ m/s}$$

The hydrogen atom recoils at a gentle speed of $0.814 	ext{ m/s}$, a beautiful consequence of the fundamental laws of conservation!

Visualized Solution

$K_{m a x} and Φ$

$E = K_{m a x} + Φ$

$E_{n} = - \frac{13.6}{n ^{2}} eV$

$Δ E = 2.55 eV$

$Δ L = L_{f} - L_{i}$

$p_{atom} = p_{photon}$

$v = \frac{E}{m c}$

\text{Final Answer}

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

The Photoelectric Connection

Decoding the Hydrogen Spectrum

The Angular Momentum Shift

The Recoil of the Atom

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kmax​ and Φ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=Kmax​+Φ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } En​=−n213.6​ eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔE=2.55 eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔL=Lf​−Li​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } patom​=pphoton​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=mcE​