A nucleus A, With a finite de-broglie wavelength `lambda_(A)`, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, While C flies in the opposite direction with a velocity equal to half of that of B . The de-Broglie wavelength `lambda_(B)` and `lambda_(B)` and C are respectively:

Options

A. `lamda_A , 2lamda_A`
B. `2lamda_A , lamda_A`
C. `lamda_A , (lamda_A)/2`
D. ` (lamda_A)/2,lamda_A `

Answer

D

Explanation

Let 2m be the mass of A and moving with velocity `v_0` There are two equal nuclei hence mass of each is m. Velocity of B is v then velocity of C will be v/2 in opposite directions as shown in diagram

According to principal of conservation of linear momentum we can write the following .
`2mv_0=mv-mv//2impliesv=4v_0`
Linear momentum of particles are as follows.
`P_A=2mv_0=P_B=4mv_0 and P_C=2mv_0`
De - Broglie wavelength `lamda =h//p`
`lamda_A =h/(2mv_0),lamda_B=h/(4mv_0)=lamda_A/2,lamda_C=h/(2mv_0)=lamda_A`