TG EAMCET 3 May 2025 Shift 2 Paper

Step 1: Displacement equationsThe proton and the alpha particle both start from rest, so their displacement after some time is given by:$s = \frac{1}{2} a t^2 .$Since they travel the same distance:$\frac{1}{2} a_\alpha t_\alpha^2 = \frac{1}{2} a_p t_p^2$This means:$a_\alpha t_\alpha^2 = a_p t_p^2$Step 2: Acceleration in an electric fieldThe acceleration of a charged particle in an electric field $E$ is:$a = \frac{q E}{m}$- For a proton, charge $= e$, mass $= m_p$- For an alpha particle, charge $= 2 e$, mass $= 4 m_p$So: $-a_p = \frac{e E}{m_p} - a_\alpha = \frac{2 e E}{4 m_p} = \frac{e E}{2 m_p}$Step 3: Substitute accelerations into equationPlug the accelerations into the earlier equation:$\left( \frac{e E}{2 m_p} \right) t_\alpha^2 = \left( \frac{e E}{m_p} \right) t_p^2$Step 4: Solve for time ratioDivide both sides by $\frac{e E}{m_p}$ :$\frac{t_\alpha^2}{2} = t_p^2$This gives: $t_\alpha^2 = 2 t_p^2$Taking square roots:$t_\alpha = \sqrt{2} t_p$So,$\frac{t_p}{t_\alpha} = \frac{1}{\sqrt{2}}$or$t_p : t_\alpha = 1 : \sqrt{2}$