Which of the following equation correctly represents the momentum of a photon of energy E

-

Question Id: PH0000003

Question: Which of the following equation correctly represents the momentum of a photon of energy E?

(a) \(\frac{E}{c}\)

(b) \(E^{2}c\)

 (c) \(Ec\)

 (d) \(Ec^{2}\)

Solution:

Here, option (a) \(\frac{E}{c}\) is the correct answer.

Explanation:

Energy carried by a photon having frequency \(\nu\) is \(E=h\nu\) ; Where \(h\) is Plank's constant.

\(\Rightarrow E=\frac{hc}{\lambda}\)

\(\Rightarrow \lambda=\frac{hc}{E}\) ; Where \(\lambda\) is wavelength and \(c\) is velocity of light.

We know the de Broglie wavelength \(\lambda=\frac{h}{p}\) 

\[\Rightarrow p=\frac{h}{\lambda}\]

\(\Rightarrow p=\frac{h}{\frac{hc}{E}}\) ; putting the value of \(\lambda\)

\[\Rightarrow p=\frac{hE}{hc}\]

\[\Rightarrow p=\frac{E}{c}\]

So, this is the equation of momentum of a photon.


 

If you have any problem regarding this please leave a comment 💬  below

Comments

Post a Comment

Popular posts from this blog

Express \(Sine(sin)\), \(Cosine(cos)\) and \(tangent(tan)\) in terms of sides of triangle

-
Image
Question : Express \(Sine(sin)\), \(Cosine(cos)\) and \(tangent(tan)\) in terms of sides of triangle  Answer : For a right angle triangle  we can find sine or sin, cosine or cos, tangent or tan in terms of their side.  To know more accurately about sin, cos and tan we must have to know about (1)base, (2)height (perpendicular) and (3)hypotenuse. Figure-1 (right angle triangle)    In general we use base, hypotenuse and height(perpendicular) as shown in figure.  But this method is not   always right.    Let's know what is right (1) Definition of hypotenuse:       The side opposite to the right angle or \(90^{o}\) is called hypotenuse, it is the largest side of right angle triangle.  (2) Definition of perpendicular:   It depends on the angle which we are going to consider. If we consider angle C the  perpendicular  is AB.  If we consider angle A then  perpendicular  is  BC. (See be...
Read more

Find the (a) maximum frequency (b) minimum wavelength of X-rays produced by 30 kV electrons.

-
Image
Question Id:  PH0000012 Question : Find the  (a) maximum frequency (b) minimum wavelength of X-rays produced by 30 kV electrons. Answer : (a) Maximum frequency of the X-ray produced by 30kV electron is \(\ 7.253 \times 10^{18}Hz \). (b) Minumum wavelength ( \(\lambda\) ) is \(0.41 \) angstroms. Explanation : Given, potential \(V=30kV\)                    \(\Rightarrow V=30 \times 10^3V\)                    \(\Rightarrow V=3 \times 10^4V\) So, energy of the X-ray is \(E=3 \times 10^4eV\) [ \(1eV=1.602\times 10^{-19}J\) ] We know a particle having frequency ( \(\nu\) ) contain energy \(E=h\nu\) ; where h is Planck's constant \(h=6.626\times 10^{-34}Js\) \[\Rightarrow \nu =\frac{E}{h}\] \[\Rightarrow \nu =\frac{1.602\times 10^{-19} \times 3 \times 10^{4}}{6.626\times 10^{-34}}\] \[\Rightarrow \nu =\frac{1.602 \times 3\times 10^{19} }{6.626}\] \[\Rightarrow \nu =0.7253 \times ...
Read more

A right angle triangle has a base of \(25m\) and its area is \(375m^{2}\) find its height

-
Image
Question : A triangle has a base of  \(25m\)  and its area is  \(375m^{2}\)  find its height?  Answer : Height of the right angle triangle is  \(30m\)  . Explanation : Given base is  \(25m\)  and area is  \(375m^{2}\) Figure-1 (right angle triangle) Now we know area of right angle triangle=  \(\frac{1}{2}\times base \times height\) So according to question,  \(\frac{1}{2}\times base \times height\) =375 \[\Rightarrow base \times height= 375 \times 2\] \[\Rightarrow 25 \times height= 750\] \[\Rightarrow height= \frac{750}{25}\] \[\Rightarrow height= 30\] Hence, height of the right angle triangle is \(30m\)  . (Answer)
Read more