Answer
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Hint: Since, the De-Broglie wavelength of a particle is inversely proportional to its momentum p. From here the wavelength is compared to the atomic structure of the hydrogen atom. Further, we can get the result by putting the values given.
Formula used:
$2\pi {r_n} = n{\lambda _n}$
Complete step by step answer:
According to wave particle duality, the De-Broglie wavelength of a particle is inversely proportional to its momentum p. The de-Broglie wavelength of a particle indicates the length scale at which wave-like properties are seen in a particle.
The atomic structure of hydrogen is given by:
$\eqalign{& {2\pi {r_n} = n{\lambda _n}} \cr}$
$\eqalign{& {\lambda _3} = \dfrac{{2\pi (4.65 \times {{10}^{ - 10}})}}{3} \cr}$
$\eqalign{& {\lambda _3} = 9.7\mathop A\limits^ \circ \cr}$
So, here r is the radius of the hydrogen atom and $\lambda $ is the de-Broglie wavelength and n is the principal quantum number.
Therefore, the correct option is C) i.e., the de-Broglie wavelength is given by $9.7\mathop A\limits^ \circ $ .
Additional Information:
Wave- particle duality is a concept of quantum mechanics that according to this every particle may be described as either a particle or a wave. It expresses the inability of the classical concepts of particle and wave, to fully describe the behavior of quantum scale objects.
In general, an electron in a metal has a de-Broglie wavelength in order of approx 10nm. So, we observe quantum-mechanical effects in the properties of a metal when the width of the sample is around this value. The S.I unit of this wavelength is meter (m).
De-Broglie won the Nobel Prize for physics in 1929, after the wave- like behavior of matter was experimentally demonstrated in 1927.
Note:
De- Broglie wavelength gives the relation between particle and wave property. If a particle is larger than its de-Broglie wavelength, or if it is interacting with other objects on a scale significantly larger than its de Broglie wavelength, then its wave- like properties are not acceptable.
Formula used:
$2\pi {r_n} = n{\lambda _n}$
Complete step by step answer:
According to wave particle duality, the De-Broglie wavelength of a particle is inversely proportional to its momentum p. The de-Broglie wavelength of a particle indicates the length scale at which wave-like properties are seen in a particle.
The atomic structure of hydrogen is given by:
$\eqalign{& {2\pi {r_n} = n{\lambda _n}} \cr}$
$\eqalign{& {\lambda _3} = \dfrac{{2\pi (4.65 \times {{10}^{ - 10}})}}{3} \cr}$
$\eqalign{& {\lambda _3} = 9.7\mathop A\limits^ \circ \cr}$
So, here r is the radius of the hydrogen atom and $\lambda $ is the de-Broglie wavelength and n is the principal quantum number.
Therefore, the correct option is C) i.e., the de-Broglie wavelength is given by $9.7\mathop A\limits^ \circ $ .
Additional Information:
Wave- particle duality is a concept of quantum mechanics that according to this every particle may be described as either a particle or a wave. It expresses the inability of the classical concepts of particle and wave, to fully describe the behavior of quantum scale objects.
In general, an electron in a metal has a de-Broglie wavelength in order of approx 10nm. So, we observe quantum-mechanical effects in the properties of a metal when the width of the sample is around this value. The S.I unit of this wavelength is meter (m).
De-Broglie won the Nobel Prize for physics in 1929, after the wave- like behavior of matter was experimentally demonstrated in 1927.
Note:
De- Broglie wavelength gives the relation between particle and wave property. If a particle is larger than its de-Broglie wavelength, or if it is interacting with other objects on a scale significantly larger than its de Broglie wavelength, then its wave- like properties are not acceptable.
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