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Particle A of mass mA=m/2 moving along t...

Particle A of mass `m_A=m/2` moving along the x - axis with velocity `v_0` collides elastically with another particle B at rest having mass `m_B=m/3` . If both particles move along the x - axis after the collision , the change `Deltalamda` in de - Broglie wavelength of particle A , in terms of its de - Broglie wavelength `(lamda_0) ` before collision is :

A

`Delta lamda= (3)/(2) lamda_(0)`

B

`Delta lamda= (5)/(2) lamda_(0)`

C

`Delta lamda = 2 lamda_(0)`

D

`Delta lamda= 4 lamda_(0)`

Text Solution

Verified by Experts

The correct Answer is:
D

Applying momentum conservation `(m)/(2) xx v_(0) + (m)/(3) xx (0)= (m)/(2) v_(A) + (m)/(3) v_(B)`
`= (v_(0))/(2) = (v_(A))/(2) + (v_(B))/(3)` ...(i)
Since, collsion is elastic (e=1)
`e=1 = (v_(B)- v_(A))/(v_(0))rArr v_(0)= v_(B)- v_(A)` ...(ii)
on solving eq (i) & (ii): `v_(A)= (v_(0))/(5)`
Now, de-Broglie wavelength of A before collision: `lamda_(0)= (h)/(m_(A) v_(0))= (h)/(((m)/(2))v_(0))`
`rArr lamda_(0)= (2h)/(mv_(0))`
Final De-Broglie wavelength
`lamda_(f)= (h)/(m_(A) v_(0))= (h)/((m)/(2) xx (v_(0))/(5)) rArr lamda_(f)= (10h)/(mv_(0))`
Now, `Delta lamda= lamda_(f)- lamda_(0)`
`Delta lamda= (10h)/(mv_(0))- (2h)/(mv_(0))`
`rArr Delta lamda= (8h)/(mv_(0)) rArr Delta lamda= 4 xx (2h)/(mv_(0))`
`rArr Delta lamda= 4lamda_(0)`
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