The radiation corresponding to 3 - 2 transition of hydrogen atom falls on a metal surface to produc

Q: Why do we use the *largest* circular path to find $KE_{\text{max}}$?

The radius $r$ is proportional to velocity $v$. Therefore, the largest radius corresponds to the fastest electrons, which possess the maximum kinetic energy ($KE_{\text{max}}$) needed for the photoelectric equation.

Q: Can I use the formula $E = -13.6/n^2$ directly?

Yes! The energy of the emitted photon is simply the difference between the initial and final energy levels: $E_3 - E_2 = (-1.51 \text{ eV}) - (-3.4 \text{ eV}) = 1.89 \text{ eV}$.

Q: Why did we convert the kinetic energy from Joules to eV?

In atomic physics, energies are typically very small, making Joules cumbersome. Since the photon energy and the options are in eV, converting $KE$ to eV ensures all terms in the photoelectric equation have consistent units.

Q: What happens if the photon energy is less than the work function?

If $h\nu < W$, no photoelectrons will be emitted, regardless of the intensity of the light. This is the threshold condition of the photoelectric effect.

Q: Why does the magnetic field not change the kinetic energy of the electron?

Magnetic force is always perpendicular to the velocity vector. It acts as a centripetal force, changing only the direction of motion, not the speed or kinetic energy.

Question

The radiation corresponding to $3 ightarrow 2$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of $3 	imes 10^{-4}$ T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to

Options:

(A) 1.8 eV

(B) 1.1 eV

(C) 0.8 eV

(D) 1.6 eV

Sigma Percentile · Accepted Answer

### The Grand Setup: Three Phenomena in One

Imagine you are witnessing a cosmic relay race, where energy is passed from one physical phenomenon to another. This problem is a beautiful synthesis of three major chapters in physics: Bohr's atomic model, Einstein's photoelectric effect, and the magnetic effects of current.

The journey begins deep inside a hydrogen atom. An electron, initially excited in the third orbit ($n=3$), decides to drop down to the second orbit ($n=2$). This quantum leap releases a burst of energy in the form of a photon. This photon doesn't just vanish; it travels through space and strikes a metal surface.

Upon impact, the photon transfers its energy to an electron bound within the metal. If the energy is sufficient, the electron breaks free, becoming a photoelectron. But the story doesn't end there. This newly freed electron is immediately subjected to a uniform magnetic field. Because it's a moving charge in a magnetic field, it experiences a Lorentz force that bends its path into a perfect circle. Our mission is to trace this energy backward from the circular path to find the metal's work function.

### Phase 1: The Hydrogen Atom's Gift

Our first task is to determine exactly how much energy the hydrogen atom gifted to the photon. According to Bohr's model, the energy of a photon emitted during a transition between two energy levels is given by the Rydberg formula:

$$E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV}$$

Here, the electron falls from the higher energy state $n_2 = 3$ to the lower energy state $n_1 = 2$. Let's substitute these quantum numbers into our master equation:

$$h
u = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) 	ext{ eV}$$

Squaring the numbers, we get:

$$h
u = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) 	ext{ eV}$$

Finding the common denominator (which is 36), the fraction simplifies to $\frac{5}{36}$. Multiplying this by 13.6 gives us the exact energy of the incident photon:

$$h
u = 1.89 	ext{ eV}$$

This $1.89 	ext{ eV}$ is the total energy budget we have for the next phase of our journey.

### Phase 2: The Magnetic Dance

Now, let's look at the end of the story. The ejected electron is moving in a circular path of radius $r = 10.0 	ext{ mm}$ inside a magnetic field $B = 3 	imes 10^{-4} 	ext{ T}$. Why does it move in a circle? Because the magnetic force provides the necessary centripetal force. The radius of this path is given by:

$$r = \frac{mv}{eB}$$

We need the kinetic energy of this electron, which is $KE_{	ext{max}} = \frac{1}{2}mv^2$. By rearranging our radius formula to solve for velocity ($v = \frac{reB}{m}$) and substituting it into the kinetic energy equation, we get a direct relationship:

$$KE_{	ext{max}} = \frac{r^2 e^2 B^2}{2m}$$

This is a powerful derivation because it allows us to find the kinetic energy directly from the radius without needing to calculate the velocity first. Now, we must carefully substitute the standard values. The radius $r = 10^{-2} 	ext{ m}$, the elementary charge $e = 1.6 	imes 10^{-19} 	ext{ C}$, the magnetic field $B = 3 	imes 10^{-4} 	ext{ T}$, and the mass of an electron $m = 9.1 	imes 10^{-31} 	ext{ kg}$.

$$KE_{	ext{max}} = \frac{(10^{-2})^2 	imes (1.6 	imes 10^{-19})^2 	imes (3 	imes 10^{-4})^2}{2 	imes 9.1 	imes 10^{-31}} 	ext{ J}$$

Squaring the terms in the numerator:

$$KE_{	ext{max}} = \frac{10^{-4} 	imes 2.56 	imes 10^{-38} 	imes 9 	imes 10^{-8}}{18.2 	imes 10^{-31}} 	ext{ J}$$

Multiplying the constants ($2.56 	imes 9 = 23.04$) and combining the powers of ten:

$$KE_{	ext{max}} = \frac{23.04 	imes 10^{-50}}{18.2 	imes 10^{-31}} 	ext{ J} \approx 1.266 	imes 10^{-19} 	ext{ J}$$

Since our photon energy is in electron-volts, we must convert this kinetic energy into eV by dividing by $1.6 	imes 10^{-19}$:

$$KE_{	ext{max}} = \frac{1.266 	imes 10^{-19}}{1.6 	imes 10^{-19}} 	ext{ eV} \approx 0.79 	ext{ eV}$$

### Phase 3: The Photoelectric Finale

We finally have all the pieces of the puzzle. We know the total energy supplied by the photon ($1.89 	ext{ eV}$) and the maximum kinetic energy the electron left with ($0.79 	ext{ eV}$). Einstein's photoelectric equation ties it all together. It states that the incident energy is split into two parts: the work function ($W$) required to break the electron free, and the remaining kinetic energy ($KE_{	ext{max}}$).

$$h
u = W + KE_{	ext{max}}$$

Rearranging to solve for the work function:

$$W = h
u - KE_{	ext{max}}$$

Substituting our calculated values:

$$W = 1.89 	ext{ eV} - 0.79 	ext{ eV}$$

$$W = 1.1 	ext{ eV}$$

And there we have it! The work function of the metal is **1.1 eV**. By tracing the physical effects backward—from the magnetic circular path to the photoelectric emission, and finally to the atomic transition—we successfully decoded the properties of the metal.

Select Answer:

Visualized Solution

The Physical Setup

Energy of Emitted Photon

Substituting Transition Values

Calculating Photon Energy

Photoelectric Equation

Electron in Magnetic Field

Kinetic Energy from Radius

Substituting Given Values

Calculating Kinetic Energy in Joules

Converting to Electron Volts

Calculating Work Function

What if the magnetic field was stronger?

The Sigma Insight: Photoelectric Effect

The Grand Setup

Phase 1

Phase 2

Phase 3

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