Explanation
Hint:
Energy mass relation for photon is given as,
$$E = m{c^2}$$
Where $$E$$ is the total energy of the photon, $$m$$ is the moving mass of the photon, and $$c$$ is the speed of light
Correct Option is A.
Explanation for correct answer:
$$\bullet $$ A photon is moving with frequency $$v$$, then its energy is given as,
$$E = hv$$
$$\bullet $$ By energy mas relation, we have
$$ \Rightarrow m{c^2} = hv$$
$$ \Rightarrow mc = \dfrac{{hv}}{c}$$ $$ \Rightarrow p = \dfrac{{hv}}{c}$$
Where $$p$$ is the momentum of the photon.
This is the required expression. Momentum of photon is given by, $$p = \dfrac{{hv}}{c}$$
Thus, Option A is correct.
The laser beams are perfectly parallel. So, the laser beams are very narrow and can travel very long distances without spreading.
$$ \lambda $$ is directly proportional to $$ \dfrac{1}{mv}$$
$$\dfrac{ \lambda_1}{ \lambda_2} = \dfrac{m_2 v_2}{ m_1 v_1}$$
$$ \dfrac{ \lambda_1}{ \lambda_2} = \dfrac{1}{2} \times \dfrac{1}{2}$$
$$ \lambda_2 = 4 \lambda_1$$
Molecular mass of CaCO3 = (40+12+16*3)=100 g/mole
No. of moles= 10/100 =0.1 mole
No. Of protons in one molecule of CaCO3=(20+6+8*3)=50
No. of protons in 0.1 mole of CaCO3 will be=(0.1*50)*(6.022*10^23)
=3.011*10^24 protons
$$\lambda \propto {1 \over m}$$
$${{{\lambda _1}} \over {{\lambda _2}}} = {{{m_2}} \over {{m_1}}} = {{3m} \over {2m}}$$
$$ = 3:2$$
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