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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