
The circumference of the second orbit of an atom or ion having a single electron, is $4 \times {10^{-9}}m$ . The de-Broglie wavelength of electron revolving in this orbit should be
A. $2 \times {10^{-9}}m$
B. $4 \times {10^{-9}}m$
C. $8 \times {10^{-9}}m$
D. $1 \times {10^{-9}}m$
Answer
556.8k+ views
Hint: The electron is given to be in the second orbit of an atom or ion. The circumference is also given.
De-Broglie waves are the property of a material that varies in time space and behaves like a wave
Formula used:
$n\lambda = 2\pi r$
Here $n$ is the number of the orbit
$\lambda $ is the de-Broglie wavelength
$r$ is the radius of the orbit.
Complete step by step answer:
The de-Broglie wavelength is a wavelength which all objects in quantum mechanics have. It helps to determine the probability density of finding the object at a given point of the space.
The circumference is given as $n\lambda $ which is given to be $4 \times {10^{-9}}m$ . But since the given circumference is of second orbit hence the value of $n = 2$ .
From above discussion, we can have:
$circumference = n\lambda $
Substituting the values of circumference from the question and value of $n = 2$ , we have:
$ \Rightarrow 4 \times {10^{-9}}m = 2\lambda $
$ \Rightarrow \lambda = \dfrac{{4 \times {{10}^{-9}}}}{2}$
$ \therefore \lambda = 2 \times {10^{ - 9}}m$
Therefore, the de-Broglie wavelength is $2 \times {10^{-9}}m$
So, the correct answer is “Option A”.
Additional Information:
It is believed that matter has the dual nature of being a particle as well as waves. de-Broglie waves are the property of a material that varies in time space and behaves like a wave.It is also known as matter-waves. It is similar to the dual nature of light.
Note:
The circumference is given as $n\lambda $ . The value of $n = 2$ is not given explicitly but mentioned in the question.
Also remember that ${10^{ - 9}}m$ is also known as nanometres.
De-Broglie wavelength is associated with an object and it is also related to its momentum and mass.
Since the electrons go in circles around the nuclei of an atom, there exists de-Broglie wavelength as a closed loop. The orbits in which the electrons circle the nucleus is known as stationary orbits.
De-Broglie waves are the property of a material that varies in time space and behaves like a wave
Formula used:
$n\lambda = 2\pi r$
Here $n$ is the number of the orbit
$\lambda $ is the de-Broglie wavelength
$r$ is the radius of the orbit.
Complete step by step answer:
The de-Broglie wavelength is a wavelength which all objects in quantum mechanics have. It helps to determine the probability density of finding the object at a given point of the space.
The circumference is given as $n\lambda $ which is given to be $4 \times {10^{-9}}m$ . But since the given circumference is of second orbit hence the value of $n = 2$ .
From above discussion, we can have:
$circumference = n\lambda $
Substituting the values of circumference from the question and value of $n = 2$ , we have:
$ \Rightarrow 4 \times {10^{-9}}m = 2\lambda $
$ \Rightarrow \lambda = \dfrac{{4 \times {{10}^{-9}}}}{2}$
$ \therefore \lambda = 2 \times {10^{ - 9}}m$
Therefore, the de-Broglie wavelength is $2 \times {10^{-9}}m$
So, the correct answer is “Option A”.
Additional Information:
It is believed that matter has the dual nature of being a particle as well as waves. de-Broglie waves are the property of a material that varies in time space and behaves like a wave.It is also known as matter-waves. It is similar to the dual nature of light.
Note:
The circumference is given as $n\lambda $ . The value of $n = 2$ is not given explicitly but mentioned in the question.
Also remember that ${10^{ - 9}}m$ is also known as nanometres.
De-Broglie wavelength is associated with an object and it is also related to its momentum and mass.
Since the electrons go in circles around the nuclei of an atom, there exists de-Broglie wavelength as a closed loop. The orbits in which the electrons circle the nucleus is known as stationary orbits.
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