Explanation
Hint: In the relativistic case the momentum is defined as the ratio of the energy and the speed of light.
Step1: Finding the transferred momentum
P=\dfrac{E}{C}
When light gets reflected perfectly, change in momentum is,
\Delta \mathrm{P}=2 \dfrac{\mathrm{E}}{\mathrm{C}}
\therefore Momentum transferred to the surface is \dfrac{\mathrm{2E}}{\mathrm{C}} (D)
HINT: ACCORDING TO DE BROGLIE THE WAVELENGTH OF AN OBJECT IS ASSOCIATED WITH THE MOMENTUM AND MASS OF THE BODY
STEP 1: Write the formula of momentum.
\mathrm{p}=\dfrac{\mathrm{h}}{\lambda}
As \lambda increases, p decreases.
Step 2: Find the correct graph.
Option(A) is wrong since the slope is a straight line but according to the above relation, the graph can not be a straight line.
Differentiating the momentum.
\dfrac{d p}{d x}=-\dfrac{h}{\lambda^{2}}
as the value is negative therefore the graph should be decreasing.
Therefore, option (D) is the correct answer.
De-broglie wavelength
\lambda =\dfrac{h}{mv}
\lambda \propto \dfrac{1}{m}
\beta particle has least mass.
So \beta particle has the longest wavelength.
Correct Answer: option (D)
Matter waves are waves associated with every moving particle and therefore are neither mechanical nor electromagnetic waves.
These waves have
\lambda = h\times p
Where \lambda is the wavelength, h is plank's constant, and p is the particle’s moment.
Therefore, matter waves are neither mechanical nor electromagnetic waves.
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