# The mass of an electron is $9.1 \times {10^{ - 31}}kg$. If its kinetic energy is $3.0 \times {10^{ - 25}}J$, calculate its wavelength.

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**Hint:**At first think about the definition and formula of kinetic energy. Kinetic energy is energy possessed by an object in motion. It is usually measured in units of joules ($J$) where one joule is equal to $1kg{m^2}/{s^2}$.

**Complete step by step answer:**

In the question, they given

Mass of the electron$ = 9.1 \times {10^{ - 31}}kg$

K.E$ = 3.0 \times {10^{ - 25}}J$

As we all know the energy formula,

$E = \dfrac{{hc}}{\lambda }$

Where E is energy, h is planck’s constant, $\lambda $is wavelength and c is speed of light.

The relation between kinetic energy and wavelength is,

$\lambda = \dfrac{h}{{{{(2mKE)}^{1/2}}}}$

Where h is Planck's constant, $h = 6.626 \times {10^{ - 34}}$, m is mass of the electron and K.E is kinetic energy.

$\lambda = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{(2 \times 9.1 \times {{10}^{ - 31}} \times 3.0 \times {{10}^{ - 25}})}}$

$\lambda = \dfrac{{6.626 \times {{10}^{ - 31}}}}{{{{(54.6 \times {{10}^{ - 56}})}^{1/2}}}}$

$\lambda = 1.2 \times {10^{ - 7}}m$

Wavelength is measured in meters.

The required wavelength is $1.2 \times {10^{ - 7}}m$.

**Additional Information:-**According to wave particle duality, the De-Broglie wavelength is a wavelength manifested in all the objects in quantum mechanics which determines the probability density of finding the object at a given point of the configuration space. The de Broglie wavelength of a particle is inversely proportional to its momentum. Matter waves are a central part of quantum mechanics, being an example of wave-particle duality. All matter exhibits wave-like behavior. It holds great similarity to the dual nature of the light which behaves as particle and wave, which has been proven experimentally. The wave properties of matter are only observable for very small objects. The thermal de-Broglie wavelength is approximately the average de-Broglie of the gas particles in an ideal gas at the specified temperature.

**Note:**

According to de-Broglie, the particles exhibit both light and wave nature. The wave nature is observed in only small particles. De-Broglie’s equation is $\lambda = \dfrac{h}{p}$ i.e. $\lambda = \dfrac{h}{{mv}}$, where $\lambda $ is de-Broglie wavelength.