A He +ion in ground state is fired towards a Hydrogen atom in the ground state and at rest. What should be the minimum kinetic energy (in eV) of He +ion so that both the single electron species may get excited?

Question

A He +ion in ground state is fired towards a Hydrogen atom in the ground state and at rest. What should be the minimum kinetic energy (in eV) of He +ion so that both the single electron species may get excited? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

4 m v=5 m v prime v prime=(4 v/5) (1/2) × 4 m v2-(1/2) × 5 m((4 v/5))2 =(10.2+40.8) eV K((1/5))=51 ⇒ K=255 eV

Q. A $H e^{+}$ ion in ground state is fired towards a Hydrogen atom in the ground state and at rest. What should be the minimum kinetic energy (in eV) of $H e^{+}$ ion so that both the single electron species may get excited?

$4 m v = 5 m v^{'}$
$v^{'} = \frac{4 v}{5}$
$\frac{1}{2} \times 4 m v^{2} - \frac{1}{2} \times 5 m (\frac{4 v}{5})^{2}$
$= (10.2 + 40.8) e V$
$K (\frac{1}{5}) = 51$
$\Rightarrow K = 255 e V$

Q. A He+ion in ground state is fired towards a Hydrogen atom in the ground state and at rest. What should be the minimum kinetic energy (in eV) of He+ion so that both the single electron species may get excited?

4mv=5mv′ v′=54v​ 21​×4mv2−21​×5m(54v​)2 =(10.2+40.8)eV K(51​)=51 ⇒K=255eV

Q. A $H e^{+}$ ion in ground state is fired towards a Hydrogen atom in the ground state and at rest. What should be the minimum kinetic energy (in eV) of $H e^{+}$ ion so that both the single electron species may get excited?

$4 m v = 5 m v^{'}$
$v^{'} = \frac{4 v}{5}$
$\frac{1}{2} \times 4 m v^{2} - \frac{1}{2} \times 5 m (\frac{4 v}{5})^{2}$
$= (10.2 + 40.8) e V$
$K (\frac{1}{5}) = 51$
$\Rightarrow K = 255 e V$