After absorbing a slowly moving neutron of mass mN(momentum ~ 0) a nucleus of mass M breaks into two nuclei of masses m1and 5m1(6m1= M + mN), respectively. If the de Broglie wavelength of the nucleus with mass m1is λ, the de Broglie wavelength of the other nucleus will be[AIEEE 2011]a)5λb)25 λc)λd)λ/5Correct answer is option 'C'. Can you explain this answer?

Question

After absorbing a slowly moving neutron of mass mN(momentum ~ 0) a nucleus of mass M breaks into two nuclei of masses m1and 5m1(6m1= M + mN), respectively. If the de Broglie wavelength of the nucleus with mass m1is &lambda;, the de Broglie wavelength of the other nucleus will be[AIEEE 2011]a)5&lambda;b)25 &lambda;c)&lambda;d)&lambda;/5Correct answer is option 'C'. Can you explain this answer?

Answer

Use conservation of momentum and lemda = h/p where p is momentum

Answer

To find the de Broglie wavelength of the nucleus with mass m1, we can use the de Broglie wavelength equation:

λ = h / p

where λ is the de Broglie wavelength, h is the Planck's constant, and p is the momentum of the particle.

Since the momentum of the neutron is approximately 0 before absorption, the total momentum after absorption must be conserved. Therefore, the momentum of the two nuclei combined must also be approximately 0.

Let p1 be the momentum of the nucleus with mass m1 and p2 be the momentum of the nucleus with mass 5m1.

Since momentum is conserved, we have:

0 = p1 + p2

We can rewrite this equation as:

p2 = -p1

Now, let's consider the relationship between momentum and mass:

p = mv

where p is the momentum, m is the mass, and v is the velocity.

Since the neutron is slowly moving, its velocity vN is much smaller than the velocities of the two nuclei. Therefore, we can approximate the velocities of the two nuclei as:

v1 ≈ vN
v2 ≈ -vN

where v1 is the velocity of the nucleus with mass m1 and v2 is the velocity of the nucleus with mass 5m1.

The momentum of a particle is also equal to the product of its mass and velocity:

p = mv

Therefore, we have:

p1 = m1 * v1 ≈ mN * vN
p2 = 5m1 * v2 ≈ -5mN * vN

Since p1 and p2 are equal in magnitude but opposite in direction, we can write:

|m1 * v1| = |-5m1 * v2|

Simplifying the equation, we have:

|m1 * vN| = |-5mN * vN|

Dividing both sides by mN, we get:

|m1 / mN * vN| = |-5 * vN|

Since vN is positive, we can drop the absolute value signs:

m1 / mN * vN = -5 * vN

Dividing both sides by vN, we get:

m1 / mN = -5

Solving for m1, we have:

m1 = -5mN

Since 6m1 = M, we can substitute -5mN for m1:

6(-5mN) = M

-30mN = M

Dividing both sides by -30, we get:

mN = -M / 30

Since mass cannot be negative, we can drop the negative sign:

mN = M / 30

Now, we can substitute this value of mN into the equation for the de Broglie wavelength:

λ = h / p

λ = h / (mN * vN)

λ = h / ((M / 30) * vN)

Simplifying, we have:

λ = 30h / (M * vN)

Therefore, the de Broglie wavelength of the nucleus with mass m1 is 30h / (M * vN).

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