The radiation corresponding to 3 - 2 transition of a hydrogen atom falls on a gold surface to gener

Q: Why do we convert the kinetic energy into electron volts (eV)?

Because the photon energy we calculated using the Rydberg formula is in eV. To use Einstein's photoelectric equation ($\phi = E_p - KE$), all terms must be in the same unit.

Q: How do we know the electron's path is circular?

When a charged particle enters a uniform magnetic field perpendicularly, the magnetic force acts as a centripetal force, causing it to move in a circular path.

Q: What does the "largest circular path" mean in the question?

The radius $r = mv/qB$ is directly proportional to velocity $v$. The "largest" radius corresponds to the maximum velocity, which in turn gives the maximum kinetic energy ($KE_{max}$) needed for the photoelectric equation.

Q: Why did we use $13.6$ in the photon energy formula?

$13.6 \text{ eV}$ is the ionization energy of hydrogen. The energy levels of a hydrogen atom are given by $-13.6/n^2 \text{ eV}$. The difference between two levels gives the formula we used.

Q: Can I use the formula $r = \sqrt{2mK}/qB$ directly?

Absolutely! That's a great shortcut. By substituting $v = \sqrt{2K/m}$ into $r = mv/qB$, you can find the kinetic energy $K$ directly without calculating velocity first. Just remember to convert $K$ to eV at the end!

Question

The radiation corresponding to $3 ightarrow 2$ transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of $5 	imes 10^{-4} 	ext{ T}$. Assume that the radius of the largest circular path followed by these electrons is $7 	ext{ mm}$, the work-function of the metal is (Take, mass of electron $= 9.1 	imes 10^{-31} 	ext{ kg}$)

Options:

(A) $1.36 	ext{ eV}$

(B) $1.88 	ext{ eV}$

(C) $0.16 	ext{ eV}$

(D) $0.82 	ext{ eV}$

Sigma Percentile · Accepted Answer

### The Grand Journey of Energy

Imagine the journey of energy in this problem. It is a beautiful sequence of physical phenomena. First, an electron in a hydrogen atom drops from the third orbit to the second, releasing a photon.

This photon then travels and strikes a gold surface, knocking out a photoelectron. Finally, this electron flies into a magnetic field, where it's forced into a circular path. Let's break this down step by step.

### Unlocking the Photon's Energy

Our first task is to find the energy of the photon emitted by the hydrogen atom. We know the Rydberg formula for energy difference. The energy of the photon is given by:

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

The transition is from the third state to the second state. So, we carefully substitute $n_1 = 2$ and $n_2 = 3$ into our equation.

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

Let's calculate this. $1/4 - 1/9$ gives us $5/36$. Multiplying this by $13.6$, we get the photon energy:

$$E_p = 13.6 \left( \frac{5}{36} ight) = 1.89 	ext{ eV}$$

Keep this value safe, we'll need it soon.

### The Magnetic Dance of the Electron

Now, let's shift our focus to the magnetic field. When a charged particle moves perpendicular to a uniform magnetic field, it takes a circular path. The radius of this path is given by:

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

From this, we can find the velocity of the electron by rearranging the formula:

$$v = \frac{qBr}{m}$$

Let's plug in the given values. The charge of an electron is $1.6 	imes 10^{-19} 	ext{ C}$. The magnetic field is $5 	imes 10^{-4} 	ext{ T}$. The radius is $7 	ext{ mm}$, which is $7 	imes 10^{-3} 	ext{ m}$. And the mass is $9.1 	imes 10^{-31} 	ext{ kg}$.

$$v = \frac{(1.6 	imes 10^{-19}) (5 	imes 10^{-4}) (7 	imes 10^{-3})}{9.1 	imes 10^{-31}} 	ext{ m/s}$$

Don't get intimidated by the powers of ten. If we carefully multiply the numbers and adjust the exponents, we find the velocity of the emitted electron:

$$v = 6.15 	imes 10^5 	ext{ m/s}$$

### Bridging the Gap: Kinetic Energy

Now that we have the velocity, we can find the maximum kinetic energy of the photoelectron. The formula is $\frac{1}{2} mv^2$. But remember, this gives the energy in Joules. To use it with our photon energy, we must convert it to electron volts by dividing by the elementary charge $e$.

$$KE = \frac{\frac{1}{2} mv^2}{e} 	ext{ (in eV)}$$

Let's substitute the mass, the velocity we just found, and the charge of the electron into our kinetic energy expression.

$$KE = \frac{9.1 	imes 10^{-31} 	imes (6.15 	imes 10^5)^2}{2 	imes 1.6 	imes 10^{-19}} 	ext{ eV}$$

Squaring the velocity and simplifying the expression, we get the kinetic energy:

$$KE = 1.075 	ext{ eV}$$

We are almost there!

### Einstein's Masterpiece: The Photoelectric Equation

This is where it all comes together. Einstein's photoelectric equation tells us that the energy of the incident photon is used to overcome the work function of the metal, and the rest becomes the kinetic energy of the electron.

$$E_p = \phi + KE_{max}$$

So, the work function $\phi$ is the photon energy minus the kinetic energy.

$$\phi = E_p - KE_{max}$$

We bring back our photon energy, $1.89 	ext{ eV}$, and subtract the kinetic energy we just calculated, $1.075 	ext{ eV}$.

$$\phi = 1.89 	ext{ eV} - 1.075 	ext{ eV}$$

Subtracting these values gives us:

$$\phi = 0.815 	ext{ eV} \approx 0.82 	ext{ eV}$$

This matches option (d) perfectly. **The work function of the metal is $0.82 	ext{ eV}$.**

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Visualized Solution

$The Physical Setup$

$Energy of Emitted Photon$

$Substituting Transition States$

$Calculating E_{p}$

$Electron in Magnetic Field$

$Substituting Values for Velocity$

$Calculating Velocity$

$Kinetic Energy of Electron$

$Substituting Values for KE$

$Calculating KE$

$Einstein’s Photoelectric Equation$

$Substituting Energies$

$Final Work Function$

$Food for Thought$

The Sigma Insight: Photoelectric Effect

The Grand Journey of Energy

Unlocking the Photon's Energy

The Magnetic Dance of the Electron

Bridging the Gap

Einstein's Masterpiece

Similar Questions

When the wavelength of radiation falling on a metal is changed from $500 nm$ to $200 nm$ , the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Physical Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of Emitted Photon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Transition States

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electron in Magnetic Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinetic Energy of Electron

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for KE

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating KE

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Einstein’s Photoelectric Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Work Function

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought