Question

a cracker of mass M at rest explodes in two parts of masses m1 and m2 with non zero velocities.find the ratio of the de Broglie wavelength of two particles. please answer these question​

Answer 1

Because the net external force on the cracker is zero, thus we can consider the initial and the final momentum of the cracker to be zero

Therefore,
Initial momentum = Final momentum

Because the cracker is at rest initially,

So,
Initial momentum = M(0) = 0

Let the velocity of the m1 mass after explosion be v1 and be v2 for mass m2.

Final momentum = m1v1 + m2v2

So,
initial momentum = Final momentum
So,
0 = m1v1 + m2v2
So,
m1v1 = -m2v2
m2v2 / m1v1 = -1


De-Broglie wavelength = h/Momentum
Wavelength of m1/Wavelength of m2
= (h/m1v1)/(h/m2v2)
= m2v2 / m1v1
= 1
Because, Wavelength ratio cannot be negative . Therefore, ratio will be 1 and not -1


Hope this helps you !

Answer 2

Answer:

The cracker explodes because of internal forces. Hence, the momentum will remain conserved. As initially it is at rest, total momentum after the explosion will also be equal to zero. Hence, the two equal parts will move with same speed in opposite directions.

Explanation:

According to de-Broglie’s hypothesis, the momentum of the particle is inversely proportional to the wavelength of the particle. Use the law of conservation of momentum to relate the momentum of both particles using de-Broglie’s hypothesis.

Formula used:

$p=\frac{h c}{\lambda}$

Here, h is Planck’s constant, c is the speed of light and λ

is the de Broglie wavelength.

The linear momentum of the particle of mass m moving with velocity v is,

$p=mv$

Complete step by step answer:

We know that, according to de-Broglie’s hypothesis, the momentum of the particle is,

$p=\frac{h c}{\lambda}$

Here, h is Planck’s constant, c is the speed of light and λ

is the de Broglie wavelength.

According to the law of conservation of momentum, the momentum of a system remains conserved.

Therefore, we can write,

$Mv=m1v1+m2v2$

Here, v is the velocity of parent particle, v1

is the velocity m1

and m1

is the velocity of m2

. Since the parent particle is at rest, the initial velocity v is zero. Therefore, the above equation becomes,

$0=m1v1+m2v2$

$m1v1=-m2v2$

Therefore, from the above equation, the momentum of the particle of mass m1  and the momentum of the particle of mass m2  is equal.

So, we can write,

$p1=p2$

$\Rightarrow \frac{h c}{\lambda_{1}}=\frac{h c}{\lambda_{2}}$

Planck’s constant h and speed of light c is constant for both particles. Therefore, the wavelength of these particles is the same. Therefore, we can write,

$\therefore \frac{\lambda_{1}}{\lambda_{2}}=1$