Question 216: Dual Nature of Matter and Radiation

Question

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The
collision is head on, and elastic. The ratio of the de-Broglie wavelengths $${\lambda _A}$$ to $${\lambda _B}$$ after the collision is:

Accepted Answer

From question, m
A
= M; m
B
= $${m \over 2}$$
u
A
= V and u
B
= 0
Let after collision velocity of A = V
1
and
velocity of B = V
2
Applying law of conservation of momentum,
mu = mv
1
+ $$\left( {{m \over 2}} ight){v_2}$$
$$ \Rightarrow $$ 24= 2v
1
+ v
2
........(i)
By law of collision
$$e = {{{v_2} - {v_1}} \over {u - 0}}$$
$$ \Rightarrow $$ u = v
2
- v
1
..........(ii)
[As collision is elastic, e = 1]
using eqns (i) and (ii)
v
1
= $${4 \over 3}$$ and v
2
= $${4 \over 3}v$$
We know, de-Broglie wavelength $$\lambda $$ = $${h \over p}$$
$$ 	herefore $$ $${{{\lambda _A}} \over {{\lambda _B}}} = {{{P_B}} \over {{P_A}}}$$ = $${{{m \over 2} 	imes {4 \over 3}u} \over {m 	imes {4 \over 3}}}$$ = 2

Answer

$${{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 3}$$

Answer

$${{{\lambda _A}} \over {{\lambda _B}}} = 2$$

Answer

$${{{\lambda _A}} \over {{\lambda _B}}} = {2 \over 3}$$

Answer

$${{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 2}$$

Dual Nature of Matter and Radiation

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths ${\lambda _A} to {\lambda _B}$ after the collision is:

Choose the correct answer: