If the total angular momentum quantum number of a nucleus (A= 200) isJ= 1, and if it is due to the rotation of the nucleus as a rigid body, find out the frequency. (in 1012rad/s)Correct answer is '2.3'. Can you explain this answer?

Question

Answer

To find the frequency of rotation of a nucleus, we need to use the formula for the angular momentum of a rotating body. The formula is given by:

L = Iω,

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Given:
Total angular momentum quantum number (J) = 1
Mass number of the nucleus (A) = 200

We know that for a rotating rigid body, the angular momentum is quantized according to the equation:

J = ℏ√(J(J+1))

where ℏ is the reduced Planck's constant (h/2π). Rearranging the equation, we get:

J(J+1) = (J/ℏ)^2

Since we are given J = 1, we can solve the equation to find the value of ℏ:

1(1+1) = (1/ℏ)^2
2 = (1/ℏ)^2
ℏ^2 = 1/2
ℏ = 1/√2

Now, we can calculate the moment of inertia (I) using the formula:

I = (2/5)MR^2

where M is the mass of the nucleus and R is the radius.

Given:
Mass number (A) = 200

Since the nucleus is assumed to be a rigid body, we can use the formula for the radius of a spherical nucleus:

R = 1.2A^(1/3)

Plugging in the values, we get:

R = 1.2(200)^(1/3) ≈ 7.82 fm

Now we can calculate the moment of inertia:

I = (2/5)(200)(1.2(200)^(1/3))^2 ≈ 1.15 x 10^4 fm^2

Finally, we can calculate the angular velocity (ω) using the formula:

L = Iω

Since we are given J = 1, we can substitute the values:

1 = (1/√2)ω
ω = √2 rad/s

To convert this to 10^12 rad/s, we divide by 10^12:

ω ≈ 2.3 x 10^12 rad/s

Therefore, the frequency of rotation of the nucleus is approximately 2.3 x 10^12 rad/s.

Top Courses for Physics

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams