Physics, 12.07.2019 02:30, dmc79765

Calculate the wavelength of an electron traveling with a speed of 3.21 × 105 m/s. calculate the wavelength of an electron traveling with a speed of 3.21 × 105 m/s. 2.27 × 10−9 m 2.06 × 10−39 m 9.63 × 1013 m 4.41 × 108 m

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answered: Katelyn1236

The wavelength of an electron traveling with a speed of 3.21 × 10⁵ m/s is A) 2.27× 10⁻⁹ m.FURTHER EXPLANATION

The solution to this problem requires the use of the de Broglie Wave Equation. The equation derived by Louis de Broglie describes the wave nature of any particle.

The equation allows the determination of a moving particle:

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

where

h is Planck's constant

m is the the mass of the particle

v is the speed of the particle.

Based on the equation, all particles have wavelike motions. Macroscopic objects, however, have very short wavelengths which make these motions undetectable, but for microscopic particles like the electrons, their wavelengths are detectable and measurable. This finding led to the determination that the frequency of electron waves are quantized meaning electrons can only exist in atoms at specific energies.

To calculate the wavelength of the given electron, use the de Broglie Wave Equation:

$\lambda \ = \frac{h}{mv}\\\lambda \ = \frac{6.626 \times 10^{-34} \frac{kg-m^2}{s}}{(9.11 \times 10^{-31} \ kg)(3.21 \times 10^5 \ \frac{m}{s})}\\\boxed {\lambda \ = 2.2658 \times 10^{-9} \ m}$

Since the given has 3 significant figures, the final answer should be:

$\boxed {\boxed {\lambda \ = 2.27 \times 10^{-9} \ m}}$

LEARN MORELearn more about Bohr's Model of the Atom Learn more about Heisenberg Uncertainty Principle Learn more about Quantum Mechanical Model

Keywords: de Broglie wave equation, Planck's constant, speed of electron

answered: justinc10

The wavelength of the electron moving at the speed of $3.21 \times {10^5}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ is $\boxed{2.27 \times {{10}^{ - 9}}\,{\text{m}}}$ or $\boxed{2.27\,{\text{nm}}}$ .

Further Explanation:

Given:

The speed at which the electron is travelling is $3.21 \times {10^5}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ .

Concept:

The wavelength of an electron moving at a particular speed is given by the De-Broglie’s equation that establishes the relation between the wavelength and the momentum of the electron.

The expression for the de-Broglie’s equation is written as.

$\boxed{\lambda = \frac{h}{{mv}}}$ …… (I)

Here, $\lambda$ is the wavelength, is the Planck’s constant, is the mass of the electron and is the velocity of the electron.

The value of the Planck’s constant is $6.626 \times {10^{ - 34}}\,{\text{J}} \cdot {\text{s}}$ and the mass of the electron is $9.1 \times {10^{ - 31}}\,{\text{kg}}$ .

Substitute the values in above expression.

$\begin{aligned}\lambda &= \frac{{6.626 \times {{10}^{ - 34}}\,{\text{J}} \cdot {\text{s}}}}{{\left( {9.1 \times {{10}^{ - 31}}\,{\text{kg}}} \right)\left( {3.21 \times {{10}^5}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}} \right)}}\\&= 2.2658 \times {10^{ - 9}}\,{\text{m}} \\&= 2.{\text{27}}\,{\text{nm}} \\\end{aligned}$

Thus, the wavelength of the electron moving at the speed of $3.21 \times {10^5}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ is $\boxed{2.27 \times {{10}^{ - 9}}\,{\text{m}}}$ or $\boxed{2.27\,{\text{nm}}}$ .

Learn More:

What is the threshold frequency ν0 of cesium? Note that 1 ev (electron volt)=1.60×10−19 j A property is a characteristic of a substance that can be observed and does not change the identity of the substance Calculate the wavelength of an electron (m = 9.11 × 10-28 g) moving at 3.66 × 106 m/s

Answer Details:

Grade: High School

Chapter: De-Broglie Hypothesis

Subject: Physics

Keywords: De-Broglie, momentum, mass of electron, wavelength, velocity of electron, 3.21x10^5 m/s, travelling at a speed.

answered: Guest

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answered: Guest

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