Answer
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Hint: Before we start addressing the problem, we need to know about the De-Broglie wavelength. The wavelength which is associated with an object in relation to its momentum and mass is known as the De-Broglie wavelength. It is an important concept in studying quantum mechanics and it deals with the matter waves.
Formula Used:
The formula for the De-Broglie wavelength is,
\[\lambda = \dfrac{h}{p}\]
Where, h is Planck’s constant and p is momentum.
Complete step by step solution:
Consider a photon of wavelength \[{\rm{4400}}\mathop {\rm{A}}\limits^{\rm{0}} \] that is passing through the vacuum. Then we need to find the effective mass and momentum of the photon. By the formula of De-Broglie wavelength, we can write,
\[\lambda = \dfrac{h}{{mv}}\]
Here, momentum \[p = mv\]
\[m = \dfrac{h}{{\lambda v}}\]
Now, substitute the value of Planck’s constant (h), velocity (v) and wavelength (\[\lambda \]) in the above equation, we get
\[m = \dfrac{{6.625 \times {{10}^{ - 34}}}}{{4400 \times {{10}^{ - 10}} \times 3 \times {{10}^8}}}\]
\[\Rightarrow v = 3 \times {10^8}\] (the speed of light in vacuum)
\[\Rightarrow m = 5 \times {10^{ - 36}}kg\]
Now, to find the momentum we have,
\[p = mv\]
\[\Rightarrow p = 5 \times {10^{ - 36}} \times 3 \times {10^8}\]
\[\therefore p = 1.5 \times {10^{ - 27}}kg - m{s^{ - 1}}\]
Therefore, the effective mass and momentum of the photon are \[5 \times {10^{ - 36}}kg,1.5 \times {10^{ - 27}}kg - m{s^{ - 1}}\].
Hence, option A is the correct answer.
Note:Wavelength is the distance between two crests or two troughs of the wave. It is a defining characteristic of a wave and it depends on the medium in which it is travelling. The wave changes its shape as it moves from one medium to another medium and the wavelength also changes. But the frequency does not change when the wave travels from one medium to another, that is it remains constant.
Formula Used:
The formula for the De-Broglie wavelength is,
\[\lambda = \dfrac{h}{p}\]
Where, h is Planck’s constant and p is momentum.
Complete step by step solution:
Consider a photon of wavelength \[{\rm{4400}}\mathop {\rm{A}}\limits^{\rm{0}} \] that is passing through the vacuum. Then we need to find the effective mass and momentum of the photon. By the formula of De-Broglie wavelength, we can write,
\[\lambda = \dfrac{h}{{mv}}\]
Here, momentum \[p = mv\]
\[m = \dfrac{h}{{\lambda v}}\]
Now, substitute the value of Planck’s constant (h), velocity (v) and wavelength (\[\lambda \]) in the above equation, we get
\[m = \dfrac{{6.625 \times {{10}^{ - 34}}}}{{4400 \times {{10}^{ - 10}} \times 3 \times {{10}^8}}}\]
\[\Rightarrow v = 3 \times {10^8}\] (the speed of light in vacuum)
\[\Rightarrow m = 5 \times {10^{ - 36}}kg\]
Now, to find the momentum we have,
\[p = mv\]
\[\Rightarrow p = 5 \times {10^{ - 36}} \times 3 \times {10^8}\]
\[\therefore p = 1.5 \times {10^{ - 27}}kg - m{s^{ - 1}}\]
Therefore, the effective mass and momentum of the photon are \[5 \times {10^{ - 36}}kg,1.5 \times {10^{ - 27}}kg - m{s^{ - 1}}\].
Hence, option A is the correct answer.
Note:Wavelength is the distance between two crests or two troughs of the wave. It is a defining characteristic of a wave and it depends on the medium in which it is travelling. The wave changes its shape as it moves from one medium to another medium and the wavelength also changes. But the frequency does not change when the wave travels from one medium to another, that is it remains constant.
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