Radiation with wavelength 6561 Å falls on a metal surface to produce photoelectrons. The electrons a

Q: Why do we assume the electron has maximum velocity ($v_{\text{max}}$) when it follows the largest circular path?

The radius of a charged particle's path in a magnetic field is given by $r = \frac{mv}{qB}$. Since mass ($m$), charge ($q$), and magnetic field ($B$) are constant, the radius $r$ is directly proportional to the velocity $v$. Therefore, the largest radius corresponds to the highest velocity, which is the maximum velocity of the emitted photoelectrons.

Q: Why did we divide the kinetic energy formula by $e$?

The standard kinetic energy formula, $\frac{1}{2}mv^2$, gives the result in Joules when using standard SI units (kg, m/s). However, in atomic physics, it's much more convenient to use electron-volts (eV). To convert from Joules to eV, you divide the value by the elementary charge $e$ ($1.6 \times 10^{-19} \text{ C}$).

Q: Where does the value $12400$ come from when calculating the photon energy?

The product of Planck's constant ($h$) and the speed of light ($c$) is approximately $1.986 \times 10^{-25} \text{ J}\cdot\text{m}$. When you convert Joules to eV and meters to Angstroms ($\text{\AA}$), the value $hc$ becomes approximately $12400 \text{ eV}\cdot\text{\AA}$ (or more precisely $12420$). It's a very handy shortcut for quick calculations.

Q: What would happen if the electron entered the magnetic field at an angle other than 90 degrees?

If the electron entered at an angle, its velocity would have two components: one parallel to the magnetic field and one perpendicular. The parallel component would remain constant, while the perpendicular component would cause circular motion. The resulting path would be a helix (spiral) rather than a simple circle.

Q: Why is the work function considered a constant property of the metal?

The work function represents the minimum energy required to break the bond between an electron and the metal lattice. This binding energy depends on the atomic structure and the specific material of the metal, making it a characteristic constant for that particular substance, regardless of the incident light.

Question

Radiation with wavelength $6561 	ext{ \AA}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 	imes 10^{-4} 	ext{ T}$. If the radius of the largest circular path followed by the electrons is $10 	ext{ mm}$, the work function of the metal is close to

Options:

(A) $0.8 	ext{ eV}$

(B) $1.1 	ext{ eV}$

(C) $1.8 	ext{ eV}$

(D) $1.6 	ext{ eV}$

Sigma Percentile · Accepted Answer

### The Setup: Light Meets Metal

Imagine a beam of light, specifically with a wavelength of $6561 	ext{ \AA}$, striking a metal surface. This light carries energy in discrete packets called photons. When these photons hit the metal, they can transfer their energy to electrons. If the energy is sufficient, it knocks the electrons right out of the metal! This is the famous photoelectric effect.

But the story doesn't end there. Once these electrons are free, they are immediately subjected to a uniform magnetic field of $3 	imes 10^{-4} 	ext{ T}$ that is perpendicular to their motion. As we know from electromagnetism, a moving charge in a perpendicular magnetic field experiences a force that causes it to move in a circular path.

### The Magnetic Dance

The magnetic force acts as the centripetal force keeping the electron in its circular orbit. The radius $r$ of this path is given by the equation:

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

We are told that the *largest* circular path has a radius of $10 	ext{ mm}$. For the radius to be at its maximum, the velocity of the electron must also be at its maximum, $v_{	ext{max}}$. Rearranging our formula to solve for this maximum velocity, we get:

$$v_{	ext{max}} = \frac{Ber}{m}$$

### Calculating the Kinetic Energy

Now that we have an expression for the maximum velocity, we can determine the maximum kinetic energy ($K_{	ext{max}}$) of these ejected electrons. The standard formula for kinetic energy is:

$$K_{	ext{max}} = \frac{1}{2}mv_{	ext{max}}^2$$

Substituting our expression for $v_{	ext{max}}$ into this equation yields:

$$K_{	ext{max}} = \frac{1}{2}m \left(\frac{Ber}{m}ight)^2 = \frac{B^2e^2r^2}{2m}$$

This gives us the kinetic energy in Joules. However, in atomic physics, it is much more convenient to work in electron-volts (eV). To convert Joules to eV, we simply divide by the elementary charge $e$:

$$K_{	ext{max}} (	ext{eV}) = \frac{B^2e^2r^2}{2m \cdot e} = \frac{B^2er^2}{2m}$$

Now, let's plug in the given values: $B = 3 	imes 10^{-4} 	ext{ T}$, $e = 1.6 	imes 10^{-19} 	ext{ C}$, $r = 10 	ext{ mm} = 10^{-2} 	ext{ m}$, and $m = 9.1 	imes 10^{-31} 	ext{ kg}$.

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

$$K_{	ext{max}} = \frac{9 	imes 1.6}{2 	imes 9.1} 	imes \frac{10^{-8} 	imes 10^{-19} 	imes 10^{-4}}{10^{-31}}$$

Notice how beautifully the powers of ten cancel out! The numerator's powers sum to $10^{-31}$, which perfectly cancels the denominator's $10^{-31}$. We are left with:

$$K_{	ext{max}} = \frac{14.4}{18.2} \approx 0.8 	ext{ eV}$$

### The Photon's Contribution

We know the kinetic energy of the electrons after they leave the metal. But how much energy did the incident light bring in the first place? The energy $E$ of a photon is calculated using Planck's equation:

$$E = \frac{hc}{\lambda}$$

A very useful shortcut when dealing with wavelengths in Angstroms is to use the approximation $hc \approx 12400 	ext{ eV}\cdot	ext{\AA}$. Using our given wavelength of $6561 	ext{ \AA}$:

$$E = \frac{12400}{6561} \approx 1.9 	ext{ eV}$$

### Einstein's Equation to the Rescue

Finally, we bring it all together using Einstein's photoelectric equation. This principle of energy conservation states that the energy of the incoming photon is used to overcome the metal's work function $\phi$ (the binding energy), with the remainder becoming the kinetic energy of the ejected electron:

$$E = \phi + K_{	ext{max}}$$

We want to find the work function $\phi$, so we rearrange the equation:

$$\phi = E - K_{	ext{max}}$$

Substituting the values we calculated:

$$\phi = 1.9 	ext{ eV} - 0.8 	ext{ eV} = 1.1 	ext{ eV}$$

**The work function of the metal is approximately $1.1 	ext{ eV}$.**

Select Answer:

Visualized Solution

Visualizing the Setup

Radius of Circular Path

Maximum Velocity

Kinetic Energy Formula

Kinetic Energy in Joules

Kinetic Energy in eV

Substituting Values

Calculating Kinetic Energy

Energy of Incident Photon

Calculating Photon Energy

Einstein's Photoelectric Equation

Calculating Work Function

The Way Forward

The Sigma Insight: Photoelectric Effect

The Setup

The Magnetic Dance

Calculating the Kinetic Energy

The Photon's Contribution

Einstein's Equation to the Rescue

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