Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

Question

Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is (A) 10.2 eV (B) 20.4 eV (C) 13.6 eV (D) 27.2 eV

Tardigrade Admin · Accepted Answer

In inelastic collision kinetic energy is not conserved so some part of

K . E

. is lost.

∴

Reduction in

K . E . =

K . E .

before collision

- K . E .

after collision
Now, since initial

K . E .

of each of two hydrogen atoms in
ground state

= 13.6 e V

∴

Total

K . E .

of both Hydrogen atom before collision

= 2 \times 13.6 = 27.2 e V

If one

H

atom goes over to first excited state

(n_{1} = 2)

and other remains in ground state

(n_{2} = 1)

then their combined

K . E .

after collision is

= \frac{13.6}{( 2 ) ^{2}} + \frac{13.6}{( 1 ) ^{2}}

= 3.4 + 13.6 = 17 e v

Hence reduced in

K . E . = 27.2 - 17

= 10.2 e V

Q. Two $H$ atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

$10.2 e V$

$20.4 e V$

$13.6 e V$

$27.2 e V$

Q. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

10.2eV

20.4eV

13.6eV

27.2eV

Q. Two $H$ atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

$10.2 e V$

$20.4 e V$

$13.6 e V$

$27.2 e V$