Question 364: Atoms & Nuclei

Question

A particle of mass 200 MeV/c
2
collides with a
hydrogen atom at rest. Soon after the collision
the particle comes to rest, and the atom
recoils and goes to its first excited state. The
initial kinetic energy of the particle (in eV) is
$${N \over 4}$$. The value of N is :
(Given the mass of the hydrogen atom to be 1
GeV/c
2
) ______ .

Accepted Answer

Given, m
particle
=  200 MeV/c
2
= m(Assume)
and m
H
= 1
GeV/c
2
= 1000 MeV/c
2
= 5 $$ 	imes $$ 200 = 5m
Applying momentum conservation,
p
i
= p
f
$$ \Rightarrow $$ mv
0
+ 0 = 0 + 5 mv'
$$ \Rightarrow $$ v' = $${{{v_0}} \over 5}$$
Initial kinetic energy, k
i
= $${1 \over 2}mv_0^2$$
Final kinetic energy, k
f
= $${1 \over 2}\left( {5m} ight){\left( {{{{v_0}} \over 5}} ight)^2}$$
$$ 	herefore $$ Loss in KE
= $${1 \over 2}mv_0^2$$ - $${1 \over 2}\left( {5m} ight){\left( {{{{v_0}} \over 5}} ight)^2}$$
= $${4 \over 5}\left( {{1 \over 2}mv_0^2} ight)$$ = $${4 \over 5}\left( {{k_i}} ight)$$
This lost energy is used by the hydrogen atom to move from ground state to the first excited state. We know the the energy required by the hydrogen atom to move from ground state to first excited state is 10.2 eV.
$$ 	herefore $$ $${4 \over 5}\left( {{k_i}} ight)$$ = 10.2
$$ \Rightarrow $$ k
i
= $${{5 	imes 10.2} \over 4} = {{51} \over 4}$$
$$ 	herefore $$ N = 51

Atoms & Nuclei

Integer / Numeric Answer Type