A light beam of wavelength 500 nm is incident on a metal having work-function of 1.25 eV, placed in

Q: Why do we assume the electrons have maximum kinetic energy when calculating the radius?

The problem states the electrons are bent into an arc of radius 30 cm. In a photoelectric experiment, electrons are emitted with a range of kinetic energies up to a maximum value. The 'maximum kinetic energy' electrons will have the largest momentum and thus the largest radius. The problem implies we are tracking these specific, most energetic electrons.

Q: Why did we convert the photon energy to eV instead of converting the work function to Joules?

You can do it either way! However, converting the photon energy to eV often leads to cleaner numbers (like 2.5 eV - 1.25 eV = 1.25 eV) which are easier to subtract mentally, reducing the chance of calculation errors.

Q: What would happen if the electron wasn't emitted exactly perpendicular to the magnetic field?

If the electron had a velocity component parallel to the magnetic field, that parallel component would remain constant (since magnetic force is zero in that direction), while the perpendicular component would cause circular motion. The combined motion would be a spiral or helix.

Q: How did we derive the formula $p = \sqrt{2mK}$?

Start with the standard kinetic energy formula: $K = \frac{1}{2}mv^2$. Multiply the numerator and denominator by mass $m$ to get $K = \frac{m^2v^2}{2m}$. Since momentum $p = mv$, we can substitute $p^2$ for $m^2v^2$, giving $K = \frac{p^2}{2m}$. Rearranging this for $p$ gives $p = \sqrt{2mK}$.

Question

A light beam of wavelength $500	ext{ nm}$ is incident on a metal having work-function of $1.25	ext{ eV}$, placed in a magnetic field of intensity $B$. The electrons emitted perpendicular to the magnetic field $B$, with maximum kinetic energy are bent into circular arc of radius $30	ext{ cm}$. The value of $B$ is ...... $	imes 10^{-7}	ext{ T}$.
(Take, $hc = 20 	imes 10^{-26}	ext{ J-m}$, mass of electron $= 9 	imes 10^{-31}	ext{ kg}$)

Sigma Percentile · Accepted Answer

### The Dance of Photons and Electrons

Imagine a microscopic game of billiards. A photon of light, acting as the cue ball, strikes a metal surface. If it has enough energy, it knocks an electron out of the metal. This is the beautiful phenomenon known as the photoelectric effect. But our story doesn't end there. Once the electron is free, it enters a region with a uniform magnetic field, which acts like an invisible hand, steering the electron into a graceful circular arc.

Let's break down this journey step-by-step to find the strength of that invisible hand—the magnetic field $B$.

### The Photoelectric Kick

First, we need to figure out how fast the electron is moving when it leaves the metal. According to Einstein's photoelectric equation, the maximum kinetic energy ($KE_{\max}$) of the emitted electron is the energy of the incoming photon minus the energy required to break free (the work function, $\phi_0$).

$$KE_{\max} = \frac{hc}{\lambda} - \phi_0$$

We are given $hc = 20 	imes 10^{-26}	ext{ J-m}$ and the wavelength $\lambda = 500 	imes 10^{-9}	ext{ m}$. Let's calculate the photon's energy in Joules:

$$E = \frac{20 	imes 10^{-26}}{500 	imes 10^{-9}} = 4 	imes 10^{-19}	ext{ J}$$

Since the work function is given in electron-volts (eV), it's much easier to convert our photon energy into eV by dividing by the elementary charge $e = 1.6 	imes 10^{-19}	ext{ C}$:

$$E = \frac{4 	imes 10^{-19}}{1.6 	imes 10^{-19}} = 2.5	ext{ eV}$$

Now, subtracting the work function $\phi_0 = 1.25	ext{ eV}$:

$$KE_{\max} = 2.5	ext{ eV} - 1.25	ext{ eV} = 1.25	ext{ eV}$$

### From Energy to Momentum

To understand how the electron bends in the magnetic field, we need its momentum $p$. The relationship between kinetic energy and momentum is a classic mechanics staple:

$$p = \sqrt{2m \cdot KE}$$

**Watch out for the trap!** We must convert the kinetic energy back to standard SI units (Joules) before plugging it into this formula.

$$KE = 1.25 	imes 1.6 	imes 10^{-19}	ext{ J} = 2.0 	imes 10^{-19}	ext{ J}$$

Now, substitute the mass of the electron $m = 9 	imes 10^{-31}	ext{ kg}$:

$$p = \sqrt{2 	imes (9 	imes 10^{-31}) 	imes (2.0 	imes 10^{-19})}$$

$$p = \sqrt{36 	imes 10^{-50}} = 6 	imes 10^{-25}	ext{ kg m/s}$$

### The Magnetic Bend

When a charged particle moves perpendicular to a uniform magnetic field, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circle. The radius $r$ of this circle is given by:

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

We know the radius $r = 30	ext{ cm} = 0.3	ext{ m}$, the momentum $p$, and the charge $e$. We just need to rearrange the formula to solve for the magnetic field $B$:

$$B = \frac{p}{er}$$

Plugging in our hard-earned numbers:

$$B = \frac{6 	imes 10^{-25}}{1.6 	imes 10^{-19} 	imes 0.3}$$

$$B = \frac{6 	imes 10^{-25}}{0.48 	imes 10^{-19}} = 12.5 	imes 10^{-6}	ext{ T}$$

The question asks for the answer in the format of $...... 	imes 10^{-7}	ext{ T}$. By shifting the decimal point, we get:

$$B = 125 	imes 10^{-7}	ext{ T}$$

And there we have it! The value we are looking for is **125**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Einstein's Photoelectric Equation

Energy of Incident Photon

Converting Energy to eV

Calculating Maximum Kinetic Energy

Relating Kinetic Energy to Momentum

Calculating Momentum

Momentum Result

Radius in Magnetic Field

Solving for Magnetic Field B

Final Calculation

Formatting the Answer

The Sigma Insight: Photoelectric Effect

The Dance of Photons and Electrons

The Photoelectric Kick

From Energy to Momentum

The Magnetic Bend

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