A stream of electrons from a heated filament was passed between two charged plates kept at a potenti

Q: Why is the kinetic energy equal to $eV$?

When a charge $e$ moves through a potential difference $V$, the electrical work done on it is $W = qV = eV$. By the work-energy theorem, if the electron starts from rest, all this work is converted into its kinetic energy.

Q: Why do we use $\frac{p^2}{2m}$ instead of $\frac{1}{2}mv^2$ for kinetic energy?

While both formulas are correct, using $\frac{p^2}{2m}$ is much more convenient here because the de Broglie wavelength formula ($\lambda = \frac{h}{p}$) directly involves momentum $p$, not velocity $v$.

Q: What does the expression $\frac{h}{\lambda}$ physically represent?

According to the de Broglie relation $\lambda = \frac{h}{p}$, rearranging it gives $p = \frac{h}{\lambda}$. Therefore, the expression $\frac{h}{\lambda}$ is simply the momentum of the electron.

Q: Does this wave-particle duality apply to macroscopic objects like a baseball?

Yes, it applies to everything! However, because Planck's constant $h$ is incredibly small, the wavelength of a massive, macroscopic object is so infinitesimally small that its wave-like properties are completely unobservable.

Q: What if the electron was already moving before it entered the potential difference?

If the electron had an initial kinetic energy $K_{initial}$, the final kinetic energy would be $K_{final} = K_{initial} + eV$. You would then use this total kinetic energy to find the final momentum and wavelength.

Question

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $h/\lambda$ (where, $\lambda$ is wavelength associated with electron wave) is given by

Options:

(A) $2 meV$

(B) $\sqrt{meV}$

(C) $\sqrt{2 meV}$

(D) $meV$

Sigma Percentile · Accepted Answer

### The Setup: Accelerating the Electron

Imagine an electron, a tiny negatively charged particle, sitting quietly at a heated filament (the cathode). Suddenly, we introduce a potential difference, $V$, between the cathode and a nearby anode. The electron feels an irresistible pull towards the positive anode and begins to accelerate, tearing across the gap.

As it moves through this electric field, the electrical work done on the electron is entirely converted into its kinetic energy. This is a beautiful demonstration of the conservation of energy. The kinetic energy $K$ gained by the electron is simply the product of its charge $e$ and the potential difference $V$:

$$K = eV$$

### The Classical Bridge: Energy and Momentum

Now, let's look at this from the perspective of classical mechanics. We know that the kinetic energy of a moving particle can also be expressed in terms of its momentum, $p$. Instead of the familiar $\frac{1}{2}mv^2$, we can write it as:

$$K = \frac{p^2}{2m}$$

This is a super useful relation because it directly connects energy to momentum, which is exactly what we need for the next step. By equating our two expressions for kinetic energy, we can find the momentum the electron has gained:

$$\frac{p^2}{2m} = eV$$

Solving for momentum $p$, we get:

$$p = \sqrt{2meV}$$

### The Quantum Leap: Matter Waves

Here comes the quantum twist! In 1924, Louis de Broglie proposed a radical idea: if light waves can act like particles (photons), then particles like electrons should be able to act like waves. He suggested that every moving particle has an associated "matter wave."

The wavelength $\lambda$ of this matter wave is inversely proportional to the particle's momentum $p$, connected by Planck's constant $h$:

$$\lambda = \frac{h}{p}$$

### Bringing It All Together

Now, we just need to synthesize our classical momentum with our quantum wavelength. Let's substitute the momentum $p$ we found earlier into the de Broglie equation:

$$\lambda = \frac{h}{\sqrt{2meV}}$$

The question specifically asks for the value of the expression $\frac{h}{\lambda}$. To find this, we simply rearrange our final equation. By multiplying both sides by $\sqrt{2meV}$ and dividing by $\lambda$, we arrive at our final answer:

$$\frac{h}{\lambda} = \sqrt{2meV}$$

This elegant result shows how the quantum properties of an electron are directly tied to the macroscopic voltage we apply to it. It perfectly matches option (c)!

Select Answer:

Visualized Solution

\text{Electron Acceleration}

\text{Kinetic Energy Gained}

\text{Kinetic Energy and Momentum}

\text{Calculating Momentum}

\text{de-Broglie Wavelength}

\text{Substituting Momentum}

\text{Final Expression}

\text{The Way Forward}

The Sigma Insight: Wave Particle Duality

The Setup

The Classical Bridge

The Quantum Leap

Bringing It All Together

Similar Questions

The de-Broglie wavelength ( $λ$ ) associated with a photoelectron varies with the frequency ( $ν$ ) of the incident radiation as, [ $ν_{0}$ is threshold frequency]

The work function of sodium metal is $4.41 \times 1 0^{- 19}$ J. If photons of wavelength $300$ nm are incident on the metal, the kinetic energy of the ejected electrons will be $(h = 6.63 \times 1 0^{- 34} J s; c = 3 \times 1 0^{8} m/s)$ ............ $\times 1 0^{- 21}$ J.

Select Answer:

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{de-Broglie Wavelength} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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

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

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