Solveeit Logo
Login

Question

Question: Compare the momentum of \({10^5}eV\) X-ray photon \(\left( {{P_x}} \right)\) with that of \({10^5}eV...

Compare the momentum of 105eV{10^5}eV X-ray photon (Px)\left( {{P_x}} \right) with that of 105eV{10^5}eV electron (Pe)\left( {{P_e}} \right)
A. PePx=12\dfrac{{{P_e}}}{{{P_x}}} = \dfrac{1}{2}
B. PePx=165\dfrac{{{P_e}}}{{{P_x}}} = \dfrac{{16}}{5}
C. PePx=15\dfrac{{{P_e}}}{{{P_x}}} = \dfrac{1}{5}
D. PePx=1\dfrac{{{P_e}}}{{{P_x}}} = 1

Explanation

Solution

Before comparing the momentum, we must know what actually momentum is. Mathematically, Momentum is the product of mass and velocity of the particle. It is a vector quantity i.e., having both magnitude and direction.

Complete step by step answer:
Firstly, finding momentum for X-ray photon:
Energy of photon be given as Eph=mphc2{E_{ph}} = {m_{ph}}{c^2}
Where c is the speed of light whose value is taken as 3×108ms−13 \times {10^8}m{s^{ - 1}}.
Mass and charge of the photon is Zero.So, we are revising the energy formula in terms of momentum,
Eph=(mphc)c{E_{ph}} = \left( {{m_{ph}}c} \right)c
⇒Eph=pphc\Rightarrow{E_{ph}} = {p_{ph}}c (∵pph=mphc)\left( {\because {p_{ph}} = {m_{ph}}c} \right)
Where pph{p_{ph}} is the momentum of the X-ray photon
Hence, momentum of photon is,
pph=Ephc ⇒pph=105×1.6×10−193×108 ⇒pph=0.53×10−22{p_{ph}} = \dfrac{{{E_{ph}}}}{c} \\\ \Rightarrow{p_{ph}}= \dfrac{{{{10}^5} \times 1.6 \times {{10}^{ - 19}}}}{{3 \times {{10}^8}}} \\\ \Rightarrow{p_{ph}}= 0.53 \times {10^{ - 22}} kgms−1kgm{s^{ - 1}}
Now finding momentum of electron,
Writing formula for energy of electron: Ee=12mev2{E_e} = \dfrac{1}{2}{m_e}{v^2}
Where me{m_e} is the mass of electron which is taken as 9.1×10−31kg9.1 \times {10^{ - 31}}kg

Revising energy formula in terms of momentum,
Ee=12(mev)2me{E_e} = \dfrac{1}{2}\dfrac{{{{\left( {{m_e}v} \right)}^2}}}{{{m_e}}}
⇒pe=mev\Rightarrow{p_e} = {m_e}v is the momentum of the electron
Hence, momentum of electron be given by,
pe=(2Eeme){p_e} = \sqrt {\left( {2{E_e}{m_e}} \right)}
⇒pe=2×105×1.6×10−19×9.1×10−31\Rightarrow {p_e}= \sqrt {2 \times {{10}^5} \times 1.6 \times {{10}^{ - 19}} \times 9.1 \times {{10}^{ - 31}}}
⇒pe=0.5396×10−22\Rightarrow{p_e} = 0.5396 \times {10^{ - 22}}
Now we just have to compare the two momentums by dividing the two momentums
So, the ratio is
∴pepph=0.539×10−220.53×10−22≈1\therefore\dfrac{{{p_e}}}{{{p_{ph}}}} = \dfrac{{0.539 \times {{10}^{ - 22}}}}{{0.53 \times {{10}^{ - 22}}}} \approx 1

Hence, option D is the correct answer.

Note: Remember mass is not the same as weight, even though we measure the mass of a body by measuring its weight, a body with the same mass weighs differently on the moon than it weighs here on earth, mass is one property of matter that doesn’t change easily.Momentum is defined as the quantity of motion of the body. It is measured by mass×velocity, as momentum depends upon velocity, and it depends on the direction of the motion of the body as well. Momentum is a vector quantity since velocity is vector while mass is scalar.