Question

A nucleus with mass number 220 initially at rest emits an alpha particle. If the Q value of reaction is 5.5 MeV, calculate the value of kinetic energy of alpha particle.

• A. 5.4 MeV
• B. 7.4 MeV
• C. 4.5 MeV
• D. 6.5 MeV

Answer 1

The Correct Option is A

Correct Answer: A

Explanation:

In a nuclear reaction, when a nucleus emits an alpha particle, the total energy released in the reaction is given by the Q value of the reaction. The Q value represents the total energy released, which is shared as kinetic energy between the products of the reaction.

In this case, the nucleus has a mass number of 220 and is initially at rest. It emits an alpha particle, and the Q value of the reaction is given as 5.5 MeV.

The energy released is shared between the alpha particle and the daughter nucleus. Since the daughter nucleus is much heavier than the alpha particle, the kinetic energy of the alpha particle will be very close to the Q value. The difference arises due to the mass ratio between the alpha particle and the daughter nucleus, but the kinetic energy of the alpha particle is approximately equal to the Q value.

Thus, the kinetic energy of the alpha particle is 5.4 MeV, which corresponds to the correct answer A.

Answer 2

Let the parent nucleus be P, the daughter nucleus be D, and the alpha particle be α.

The decay process is: P → D + α

We are given:

• Mass number of the parent nucleus, A_P = 220
• The parent nucleus is initially at rest.
• The Q-value of the reaction, Q = 5.5 MeV.

An alpha particle (α) is a Helium nucleus (⁴He), so its mass number is A_α = 4.

By conservation of mass number:

A_P = A_D + A_α

220 = A_D + 4

A_D = 220 - 4 = 216

Since the parent nucleus is at rest, the initial momentum is zero. By conservation of momentum, the total final momentum must also be zero. This means the momentum of the daughter nucleus (p_D) and the momentum of the alpha particle (p_α) are equal in magnitude and opposite in direction:

|p_D| = |p_α| = p

The Q-value is the total kinetic energy released in the decay, which is shared between the daughter nucleus and the alpha particle:

Q = K_D + K_α

Where K_D is the kinetic energy of the daughter nucleus and K_α is the kinetic energy of the alpha particle.

The kinetic energy K is related to momentum p and mass m by K = p² / (2m). Therefore:

K_D = p² / (2m_D)

K_α = p² / (2m_α)

From these equations, we can write:

p² = 2m_DK_D = 2m_αK_α

So, K_D = (m_α / m_D) * K_α

Substitute this into the Q-value equation:

Q = (m_α / m_D) * K_α + K_α

Q = K_α * ( (m_α / m_D) + 1 )

Q = K_α * ( (m_α + m_D) / m_D )

Solving for K_α:

K_α = Q * ( m_D / (m_α + m_D) )

We can approximate the ratio of masses by the ratio of their mass numbers (A):

m_D ≈ A_D = 216

m_α ≈ A_α = 4

m_α + m_D ≈ A_α + A_D = A_P = 220

So, the formula becomes:

K_α ≈ Q * ( A_D / A_P )

K_α ≈ Q * ( (A_P - A_α) / A_P )

Plugging in the values:

K_α ≈ 5.5 MeV * ( (220 - 4) / 220 )

K_α ≈ 5.5 MeV * ( 216 / 220 )

K_α ≈ 5.5 MeV * ( 54 / 55 ) (Dividing 216 and 220 by 4)

K_α ≈ (5.5 / 55) * 54 MeV

K_α ≈ 0.1 * 54 MeV

K_α ≈ 5.4 MeV

Answer: The kinetic energy of the alpha particle is approximately 5.4 MeV. This corresponds to option (A).