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GATE 2026 PH – Question 56

Classical Mechanics · central force motion: Kepler problem · 2 marks · Multiple select

An $\alpha$ particle moves towards a fixed nucleus carrying charge $Ze$, with initial speed $v_0$ and impact parameter $b$. Starting from a large distance from the nucleus, its distance of closest approach is $r_m$ and its speed there is $v_m$. Then which of the following options is/are correct? ($k=1/(4\pi\epsilon_0)$ and $r_0=kZe^2/(mv_0^2)$)

  1. $v_0b=v_mr_m$
  2. $v_0b=2v_mr_0$
  3. For $b/r_0\ll1$, $r_m=4r_0+b^2/(2r_0)$ ignoring higher order corrections in $b/r_0$
  4. For $b/r_0\ll1$, $r_m=4r_0+b^2/(8r_0)$ ignoring higher order corrections in $b/r_0$

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Correct answer: (A) $v_0b=v_mr_m$

Explanation

**Conservation laws for the α particle (charge $2e$) approaching a nucleus (charge $Ze$).**

**Angular momentum is conserved** (the force is central). At a large distance the angular momentum is $mv_0b$, and at the closest approach, where the velocity is perpendicular to the radius, it is $mv_mr_m$:
$$v_0b=v_mr_m .\quad(\text{statement A is correct}) ✓$$

**Energy conservation** with the potential energy $\dfrac{k(2e)(Ze)}{r}=\dfrac{2kZe^2}{r}$:
$$\tfrac12mv_0^2=\tfrac12mv_m^2+\frac{2kZe^2}{r_m}.$$

Using $v_m=v_0b/r_m$ and $r_0=\dfrac{kZe^2}{mv_0^2}$:
$$1-\frac{b^2}{r_m^2}=\frac{4r_0}{r_m}\;\Rightarrow\;r_m^2-4r_0r_m-b^2=0\;\Rightarrow\;r_m=2r_0+\sqrt{4r_0^2+b^2}.$$

For $b\ll r_0$:
$$r_m\approx4r_0+\frac{b^2}{4r_0},$$
which does not agree with the coefficient of $b^2/r_0$ in either option C ($\frac12$) or D ($\frac18$) as printed. Statement B ($v_0b=2v_mr_0$) is also wrong.

Only statement **A** is correct.

Official GATE 2026 answer key: https://gate2026.iitg.ac.in/doc/download/2026/Keys/PH_Keys.pdf