An electron in a hydrogen like atom is in an excited state. It has a total energy of -3.4 eV. Calcul

Q: Why is the kinetic energy positive when the total energy is negative?

In a bound system like an atom, the potential energy is negative and its magnitude is twice the kinetic energy. The total energy is the sum of kinetic and potential energy, resulting in a negative value. Kinetic energy, being related to motion ($1/2 mv^2$), must always be positive.

Q: Can we use the formula $\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA}$ here?

Yes! Since the kinetic energy is $3.4 \text{ eV}$, it is equivalent to an electron accelerated through a potential difference of $3.4 \text{ V}$. Using $\lambda = \frac{12.27}{\sqrt{3.4}} \text{ \AA}$ gives approximately $6.65 \text{ \AA}$, which is very close to the exact calculated value.

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

The formula $\lambda = \frac{h}{\sqrt{2mK}}$ requires all values to be in standard SI units to yield the wavelength in meters. Since Planck's constant and mass are in SI units, kinetic energy must be in Joules.

Q: What does the de-Broglie wavelength signify in an atom?

It represents the wave-like nature of the electron. For an electron in a stable orbit, the circumference of the orbit must be an integer multiple of its de-Broglie wavelength, forming a standing wave.

Q: Why did we use $h = 6.6 \times 10^{-34} \text{ J s}$ instead of $6.626 \times 10^{-34} \text{ J s}$?

In many competitive exams like JEE, using $6.6 \times 10^{-34} \text{ J s}$ simplifies the manual calculations and matches the provided options or answer keys perfectly. Always look at the given constants in the exam instructions.

Question

An electron in a hydrogen like atom is in an excited state. It has a total energy of $-3.4	ext{ eV}$. Calculate
(a) the kinetic energy,
(b) the de-Broglie wavelength of the electron.

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine an electron orbiting the nucleus of a hydrogen-like atom. It's not just a particle moving in a circle; it's a quantum entity with both particle and wave characteristics. We are given that the total energy of this electron in its excited state is $-3.4	ext{ eV}$. Our goal is to uncover two fundamental properties: its kinetic energy and its de-Broglie wavelength.

### The Energy Relations
In the Bohr model of a hydrogen-like atom, the electron is bound to the nucleus by the electrostatic force of attraction. This bound state means the total energy $E$ is negative. The total energy is the sum of the kinetic energy $K$ and the potential energy $U$.

A beautiful symmetry exists in these bound orbits, dictated by the Virial Theorem:
$$K = -E$$
$$U = 2E$$

Since we know the total energy $E = -3.4	ext{ eV}$, finding the kinetic energy is straightforward. We simply take the negative of the total energy:
$$K = -(-3.4	ext{ eV}) = 3.4	ext{ eV}$$

This makes perfect physical sense—kinetic energy, being related to the square of velocity, must always be a positive quantity.

### The Wave Nature of the Electron
Now, let's delve into the quantum realm. According to Louis de Broglie, every moving particle has an associated wave. The wavelength $\lambda$ of this matter wave is inversely proportional to the particle's momentum $p$:
$$\lambda = \frac{h}{p}$$

We can express the momentum $p$ in terms of the kinetic energy $K$ using the classical relation $p = \sqrt{2mK}$. Substituting this into the de-Broglie equation gives us our master formula:
$$\lambda = \frac{h}{\sqrt{2mK}}$$

### Final Calculation
Before we plug in the numbers, we must ensure all our units are consistent. We need to convert the kinetic energy from electron-volts (eV) to Joules (J).
$$K = 3.4	ext{ eV} = 3.4 	imes 1.6 	imes 10^{-19}	ext{ J}$$

Now, let's substitute the standard constants: Planck's constant $h = 6.6 	imes 10^{-34}	ext{ J s}$ and the mass of an electron $m = 9.1 	imes 10^{-31}	ext{ kg}$.
$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 3.4 	imes 1.6 	imes 10^{-19}}}$$

Let's carefully evaluate the term inside the square root:
$$2 	imes 9.1 	imes 3.4 	imes 1.6 = 99.008$$
$$\sqrt{99.008 	imes 10^{-50}} \approx 9.95 	imes 10^{-25}$$

Finally, dividing Planck's constant by this momentum yields the wavelength:
$$\lambda = \frac{6.6 	imes 10^{-34}}{9.95 	imes 10^{-25}} \approx 6.63 	imes 10^{-10}	ext{ m}$$

To express this in a more convenient unit for atomic scales, we convert meters to Angstroms ($1	ext{ \AA} = 10^{-10}	ext{ m}$):
$$\lambda = 6.63	ext{ \AA}$$

And there we have it! The electron dances around the nucleus with a kinetic energy of $3.4	ext{ eV}$, tracing out a matter wave with a wavelength of $6.63	ext{ \AA}$.

Animated Solution for Physics - Atoms and Nuclei: An electron in a hydrogen like atom is in an excited state. It has a total energy of $- 3.4 eV$ . Calculate (a) the kinetic energy, (b) the de-Broglie wavelength of the electron.

Visualized Solution

Energy Relations in Bohr Model

Calculating Kinetic Energy

De-Broglie Wavelength Formula

Substituting Values

Final Calculation

The Sigma Insight: Bohr's Atomic Model and Energy Levels

Analyzing the Setup Imagine an electron orbiting the nucleus of a hydrogen-like atom

The Energy Relations In the Bohr model of a hydrogen-like atom, the electron is bound to the nucleus by the electrostatic force of attraction

The Wave Nature of the Electron Now, let's delve into the quantum realm

Final Calculation Before we plug in the numbers, we must ensure all our units are consistent

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