
The de-Broglie’s wavelength of electrons present in the first Bohr orbit of an H atom is?
A. $4 \times 0.529{\text{ }}\mathop {\text{A}}\limits^{\text{o}} $
B. $2\pi \times 0.529{\text{ }}\mathop {\text{A}}\limits^{\text{o}} $
C. $\dfrac{{0.529}}{{2\pi }}{\text{ }}\mathop {\text{A}}\limits^{\text{o}} $
D. $0.529{\text{ }}\mathop {\text{A}}\limits^{\text{o}} $
Answer
568.2k+ views
Hint: To answer this question, you must recall the formula for de Broglie’s wavelength of an electron. De Broglie suggested that every form of matter behaves like waves in some circumstances.
Formula used: $\lambda = \dfrac{{\text{h}}}{{{\text{mv}}}} = \dfrac{{\text{h}}}{{\text{p}}}$
Where, $\lambda $is the wavelength of the matter wave
$h$ is Planck’s constant
$m$ is the mass of the particle under consideration
$v$ is the velocity of the particle under consideration
And p is the momentum of the particle
Complete step by step solution:
We know that the angular momentum of an electron moving in a circular orbit is given by
${\text{mvr}} = \dfrac{{{\text{nh}}}}{{2\pi }}$
Substituting the value of momentum from the de Broglie equation, ${\text{mv}} = \dfrac{{\text{h}}}{\lambda }$
We get, $\dfrac{{\text{h}}}{\lambda } = \dfrac{{{\text{nh}}}}{{2\pi {\text{r}}}}$
Rearranging the equation to get wavelength, we can write
$\lambda = \dfrac{{2\pi {\text{r}}}}{{\text{n}}}$
We are given that the electron under consideration is in the first orbit of the hydrogen atom.
The radius of the first orbit of hydrogen atom is $0.529{\text{ }}\mathop {\text{A}}\limits^{\text{o}} $
Also, for the first orbit, $n = 1$
Thus, $\lambda = 2\pi \times 0.529$ Å
The correct option is B.
Note: Matter waves are an essential part of the theory of quantum mechanics, being an example of dual nature of matter. It was suggested that all matter exhibits a wave-like behaviour. For example, a beam of electrons can be diffracted in the same way like a beam of light or a water wave does. In most cases, the wavelength of objects is too small to have a practical impact on our day-to-day activities. Hence in our day-to-day lives, with objects of the size of tennis balls and people, matter waves are not of significant wavelength. These matter waves are referred to as de Broglie waves.
Formula used: $\lambda = \dfrac{{\text{h}}}{{{\text{mv}}}} = \dfrac{{\text{h}}}{{\text{p}}}$
Where, $\lambda $is the wavelength of the matter wave
$h$ is Planck’s constant
$m$ is the mass of the particle under consideration
$v$ is the velocity of the particle under consideration
And p is the momentum of the particle
Complete step by step solution:
We know that the angular momentum of an electron moving in a circular orbit is given by
${\text{mvr}} = \dfrac{{{\text{nh}}}}{{2\pi }}$
Substituting the value of momentum from the de Broglie equation, ${\text{mv}} = \dfrac{{\text{h}}}{\lambda }$
We get, $\dfrac{{\text{h}}}{\lambda } = \dfrac{{{\text{nh}}}}{{2\pi {\text{r}}}}$
Rearranging the equation to get wavelength, we can write
$\lambda = \dfrac{{2\pi {\text{r}}}}{{\text{n}}}$
We are given that the electron under consideration is in the first orbit of the hydrogen atom.
The radius of the first orbit of hydrogen atom is $0.529{\text{ }}\mathop {\text{A}}\limits^{\text{o}} $
Also, for the first orbit, $n = 1$
Thus, $\lambda = 2\pi \times 0.529$ Å
The correct option is B.
Note: Matter waves are an essential part of the theory of quantum mechanics, being an example of dual nature of matter. It was suggested that all matter exhibits a wave-like behaviour. For example, a beam of electrons can be diffracted in the same way like a beam of light or a water wave does. In most cases, the wavelength of objects is too small to have a practical impact on our day-to-day activities. Hence in our day-to-day lives, with objects of the size of tennis balls and people, matter waves are not of significant wavelength. These matter waves are referred to as de Broglie waves.
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