What is the wavelength of an electron with a mass of $9\times {{10}^{-31}}kg$ and a speed of $2.5\times {{10}^{6}}m{{s}^{-1}}$.
Answer
575.1k+ views
Hint: An electron with given mass is made to move at a very high speed. The de Broglie wavelength gives the relation between the wavelength and kinetic energy of a body. The kinetic energy depends on the mass and square of velocity. Substituting given values we can calculate the de Broglie wavelength of the electron. Convert the units as required.
Formulas used:
$\lambda =\dfrac{h}{\sqrt{2mE}}$
Complete step-by-step solution:
Electrons are negatively charged particles which revolve around the nucleus in an atom. Electrons give out matter waves due to its wave nature and the wavelength of its matter waves is equal to the De-Broglie wavelength. Given, an electron has mass $9\times {{10}^{-31}}kg$ and speed $2.5\times {{10}^{6}}m{{s}^{-1}}$.
The relation between wavelength of electron and kinetic energy is given by De-Broglie's equation to calculate wavelength. Therefore,
$\lambda =\dfrac{h}{\sqrt{2mE}}$ - (1)
Here, $\lambda $ is the wavelength
$h$ is the Planck’s constant
$m$ is the mass
$E$ is the kinetic energy
We know that,
$E=\dfrac{1}{2}m{{v}^{2}}$
Here, $v$ is the velocity
To calculate the kinetic energy of electron, w substitute given values in the above equation to get,
$\begin{align}
& E=\dfrac{1}{2}\times 9\times {{10}^{-31}}\times kg\times {{(2.5\times {{10}^{6}}m{{s}^{-1}})}^{2}} \\
& \Rightarrow E=2.84\times {{10}^{-18}}J \\
\end{align}$
Therefore, the kinetic energy of the electron is $2.84\times {{10}^{-18}}J$.
Now, substituting given values in eq (1), we get,
$\begin{align}
& \lambda =\dfrac{6.626\times {{10}^{-34}}}{\sqrt{2\times 9\times {{10}^{-28}}kg\times 2.84\times {{10}^{-18}}}} \\
& \Rightarrow \lambda =9.2\times {{10}^{-12}}m \\
\end{align}$
Therefore, the wavelength of the electron is $9.2\times {{10}^{-12}}m$.
Note: For photons the wavelength is equal to Planck’s constant divided by its momentum. The De-Broglie wavelength is used to represent the wave nature of matter. All matter shows waves as well as particle-like behavior. Wavelengths associated with objects with significant sizes have small wavelengths which are irrelevant in our day to day lives.
Formulas used:
$\lambda =\dfrac{h}{\sqrt{2mE}}$
Complete step-by-step solution:
Electrons are negatively charged particles which revolve around the nucleus in an atom. Electrons give out matter waves due to its wave nature and the wavelength of its matter waves is equal to the De-Broglie wavelength. Given, an electron has mass $9\times {{10}^{-31}}kg$ and speed $2.5\times {{10}^{6}}m{{s}^{-1}}$.
The relation between wavelength of electron and kinetic energy is given by De-Broglie's equation to calculate wavelength. Therefore,
$\lambda =\dfrac{h}{\sqrt{2mE}}$ - (1)
Here, $\lambda $ is the wavelength
$h$ is the Planck’s constant
$m$ is the mass
$E$ is the kinetic energy
We know that,
$E=\dfrac{1}{2}m{{v}^{2}}$
Here, $v$ is the velocity
To calculate the kinetic energy of electron, w substitute given values in the above equation to get,
$\begin{align}
& E=\dfrac{1}{2}\times 9\times {{10}^{-31}}\times kg\times {{(2.5\times {{10}^{6}}m{{s}^{-1}})}^{2}} \\
& \Rightarrow E=2.84\times {{10}^{-18}}J \\
\end{align}$
Therefore, the kinetic energy of the electron is $2.84\times {{10}^{-18}}J$.
Now, substituting given values in eq (1), we get,
$\begin{align}
& \lambda =\dfrac{6.626\times {{10}^{-34}}}{\sqrt{2\times 9\times {{10}^{-28}}kg\times 2.84\times {{10}^{-18}}}} \\
& \Rightarrow \lambda =9.2\times {{10}^{-12}}m \\
\end{align}$
Therefore, the wavelength of the electron is $9.2\times {{10}^{-12}}m$.
Note: For photons the wavelength is equal to Planck’s constant divided by its momentum. The De-Broglie wavelength is used to represent the wave nature of matter. All matter shows waves as well as particle-like behavior. Wavelengths associated with objects with significant sizes have small wavelengths which are irrelevant in our day to day lives.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

The chemical formula of tear gas is A CO Cl 2 B C 10 class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

