Question 184: Dual Nature of Matter and Radiation

Question

Particle A of mass mA = $${m \over 2}$$ moving along the x-axis with velocity v0 collides elastically with another particle B at rest having mass mB = $${m \over 3}$$. If both particles move along the x-axis after the collision, the change $$\Delta $$$$\lambda $$ in de-Broglie wavlength of particle A, in terms of its de-Broglie wavelength ($$\lambda $$0) before collision is :

Accepted Answer

Applying momentum conservation
$${m \over 2} 	imes {V_0} + {m \over 3} 	imes 0 = {m \over 2}{V_A} + {m \over 3}{V_B}$$
$$ \Rightarrow $$ $${{{V_0}} \over 2} = {{{V_A}} \over 2} + {{{V_B}} \over 3}$$ .....(1)
Since, collision is elastic (e = 1)
e = 1 = $${{{V_B} - {V_A}} \over {{V_0}}}$$
$$ \Rightarrow $$ V
0
= V
B
Ã¢ÂÂ V
A
.....(2)
On solving (1) & (2) :
V
A
= $${{{V_0}} \over 5}$$
Now, De-Broglie wavelength of A before collision :
$$\lambda $$
0
= $${h \over {{m_A}{V_0}}}$$
= $${h \over {\left( {{m \over 2}} ight){V_0}}}$$
= $${{2h} \over {m{V_0}}}$$
Final De-Broglie wavelength :
$$\lambda $$
f
= $${h \over {{m_A}{V_A}}}$$
= $${h \over {\left( {{m \over 2}} ight)\left( {{{{V_0}} \over 5}} ight)}}$$ = $${{10h} \over {m{V_0}}}$$
Now, $$\Delta $$$$\lambda $$ = $$\lambda $$
f
- $$\lambda $$
0
= $${{10h} \over {m{V_0}}}$$ - $${{2h} \over {m{V_0}}}$$
$$ \Rightarrow $$ $$\Delta $$$$\lambda $$ = $${{8h} \over {m{V_0}}}$$
$$ \Rightarrow $$ $$\Delta $$$$\lambda $$ = $$4 	imes {{2h} \over {m{V_0}}}$$
$$ \Rightarrow $$ $$\Delta $$$$\lambda $$ = 4$$\lambda $$
0

Answer

$$\Delta $$$$\lambda $$ = $${5 \over 2}{\lambda _0}$$

Answer

$$\Delta $$$$\lambda $$ = $${3 \over 2}{\lambda _0}$$

Answer

$$\Delta $$$$\lambda $$ = 2$$\lambda $$0

Answer

$$\Delta $$$$\lambda $$ = 4$$\lambda $$0

Dual Nature of Matter and Radiation

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