
The formula of the kinetic mass of a photon is? Where h is Planck’s constant,$\upsilon $ is the frequency of the photon and c is the speed.
(A) $\dfrac{{h\upsilon }}{c}$
(B) $\dfrac{{h\upsilon }}{{{c^2}}}$
(C) $\dfrac{{hc}}{\upsilon }$
(D) $\dfrac{{{c^2}}}{{h\upsilon }}$
Answer
141k+ views
Hint The mass and energy are interchangeable according to Einstein’s special theory of relativity. The equation is given as $E = m{c^2}$ where, $m$ is the mass and $c$ is the speed of light. And the equation connecting Planck’s constant and the energy is the Planck’s equation of energy which is given as $E = h\upsilon $ , Where $h$ is the Planck’s constant and $\upsilon $ is the frequency of the photon.
Complete Step by step solution
Take Einstein’s equation of special relativity,
$E = m{c^2}$ Where, $m$ is the mass and $c$ is the speed of light.
Also consider the Planck’s equation of energy which is given as $E = h\upsilon $ , Where $h$ is the Planck’s constant and $\upsilon $ is the frequency of the photon.
Combining these two equations, we get
$
m{c^2} = h\upsilon \\
m = \dfrac{{h\upsilon }}{{{c^2}}} \\
$
We know that the proton has no rest mass. But the effective mass in the above expression says that the effective mass varies according to the frequency of the photon. Also the above expression can be explained in the particle nature that each photon is having mass $m = \dfrac{{h\upsilon }}{{{c^2}}}$are travelling at the speed of light. The photon having higher frequency and lower wavelength will have higher effective mass which implies that it would have higher energy. This is according to the mass energy conversion described in Einstein's special theory of Relativity.
The correct answer is Option B.
Note Since the protons have no mass, the term “Kinetic mass” relates to the kinetic energy of the photon. The energy of the photon is given as $E = \dfrac{{hc}}{\lambda }$ where $\lambda $ is the wavelength of the photon.
Comparing with the Einstein’s special theory of relativity,
$
m{c^2} = \dfrac{{hc}}{\lambda } \\
m = \dfrac{h}{{c\lambda }} \\
$
This is also the kinetic mass of the photon.
Complete Step by step solution
Take Einstein’s equation of special relativity,
$E = m{c^2}$ Where, $m$ is the mass and $c$ is the speed of light.
Also consider the Planck’s equation of energy which is given as $E = h\upsilon $ , Where $h$ is the Planck’s constant and $\upsilon $ is the frequency of the photon.
Combining these two equations, we get
$
m{c^2} = h\upsilon \\
m = \dfrac{{h\upsilon }}{{{c^2}}} \\
$
We know that the proton has no rest mass. But the effective mass in the above expression says that the effective mass varies according to the frequency of the photon. Also the above expression can be explained in the particle nature that each photon is having mass $m = \dfrac{{h\upsilon }}{{{c^2}}}$are travelling at the speed of light. The photon having higher frequency and lower wavelength will have higher effective mass which implies that it would have higher energy. This is according to the mass energy conversion described in Einstein's special theory of Relativity.
The correct answer is Option B.
Note Since the protons have no mass, the term “Kinetic mass” relates to the kinetic energy of the photon. The energy of the photon is given as $E = \dfrac{{hc}}{\lambda }$ where $\lambda $ is the wavelength of the photon.
Comparing with the Einstein’s special theory of relativity,
$
m{c^2} = \dfrac{{hc}}{\lambda } \\
m = \dfrac{h}{{c\lambda }} \\
$
This is also the kinetic mass of the photon.
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