JEE MAIN - Physics (2021 - 31st August Morning Shift - No. 17)
A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :
Kp < Ke and Pp = Pe
Kp = Ke and Pp = Pe
Kp < Ke an Pp < Pe
Kp > Ke and Pp = Pe
Explanation
$${\lambda _p} = {h \over {{P_p}}}$$
$${\lambda _e} = {h \over {{P_e}}}$$
$$\because$$ $${\lambda _p} = {\lambda _e}$$
$$ \Rightarrow {P_p} = {P_e}$$
$${(K)_p} = {{P_p^2} \over {2{m_p}}}$$
$${(K)_e} = {{P_e^2} \over {2{m_e}}}$$
Kp < Ke as mp > me
Option (a)
$${\lambda _e} = {h \over {{P_e}}}$$
$$\because$$ $${\lambda _p} = {\lambda _e}$$
$$ \Rightarrow {P_p} = {P_e}$$
$${(K)_p} = {{P_p^2} \over {2{m_p}}}$$
$${(K)_e} = {{P_e^2} \over {2{m_e}}}$$
Kp < Ke as mp > me
Option (a)
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