Question

Calculate the wavelength of an electron (m = 9.11 x 10-28 g) moving at 3.66 x 106 m/s.
a. 1.99 x 10-10 m
b. 5.03 x 10-10 m
c. 1.81 x 10-10 m
d. 5.52 x 10-9 m
e. 2.76 x 10-9 m

Answer 1

To answer this question, we will use De Broglie's rule which states that:
lambda = Planck's constant / momentum
Since: momentum = mass * velocity
Therefore,
lambda = Planck's constant / (mass*velocity)

For this problem, we are give that:
h (Planck's constant) = 6.63 * 10^-34
mass = 9.11 * 10^-28 * 10^-3 kg
velocity = 3.66 * 10^6 meter/sec

Substitute in De Brouglie's formula to get the wavelength as follows:
lambda = (6.63*10^-34) / (9.11*10^-28*10^-3*3.66*10^6)
lambda = 1.99 * 10^-10 meters

Correct choice is A

Answer 2

The wavelength of an electron is $\boxed{{\text{a}}.1.99\times{{10}^{-10}}{\text{m}}}$

Further Explanation:

de Broglie wavelength:

The de Broglie equation is used to calculate the wavelength from the given values of mass and velocity. It is specially applied to neutral atoms, elementary particles, and molecules. The de Broglie equation is as follows:

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

Here, m is the mass in kilogram, h is the Planck’s constant whose value is equal to $6.626\times{10^{-34}}{\text{ J}}\cdot {\text{sec}}$, λ is the de Broglie wavelength in meters, and v is the velocity in meter per second.

The velocity of a given electron is $3.66\times{10^6}{\text{ m/s}}$.

The mass of a given electron is $9.11\times{10^{-28}}\,{\text{g}}$.

Mass of an electron in kilogram is calculated as follows:

$\begin{aligned}{\text{Mass}}\left({{\text{kg}}}\right)&={\text{Mass}}\left({\text{g}}\right)\times\left( {\frac{{1{\text{kg}}}}{{1000{\text{ g}}}}}\right)\\&=\left({9.11\times{{10}^{-28}}\,{\text{g}}}\right)\times\left({\frac{{1{\text{ kg}}}}{{1000{\text{ g}}}}}\right)\\&=9.11\times{10^{-31}}\,{\text{kg}}\\\end{aligned}$

Substitute $6.626\times{10^{-34}}{\text{J}}\cdot {\text{sec}}$ for h, $9.11\times{10^{-31}}\,{\text{kg}}$ for m, and $3.66\times{10^6}{\text{m/s}}$ for v in equation (1) to calculate the value of de Broglie wavelength (λ).

$\begin{aligned}\lambda&=\frac{h}{{mv}}\\&=\frac{{\left( {6.626\times{{10}^{-34}}{\text{ J}}\cdot {\text{sec}}}\right)}}{{\left({9.11 \times{{10}^{-31}}\,{\text{kg}}}\right)\left({3.66\times{{10}^6}{\text{m/s}}}\right)}}\\&=1.9872\times{10^{-10}}{\text{m}}\\&\approx1.99\times{10^{-10}}{\text{m}}\\\end{aligned}$

Learn more:

1. Balanced chemical equation: https://brainly.com/question/1405182

2. Identification of all of the phases of the reaction: https://brainly.com/question/8926688

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: de Broglie equation, m, h, v, wavelength, electron, velocity, mass, planck’s constant, kilogram and 1.99*10-10 m.