In the Bohr model of a `pi-mesic` atom , a `pi-mesic` of mass `m_(pi)` and of the same charge as the electron is in a circular orbit of ratio of radius `r` about the nucleus with an orbital angular momentum `h//2 pi`. If the radius of a nucleus of atomic number `Z` is given by `R = 1.6 xx 10^(-15) Z^((1)/(3)) m`, then the limit on `Z` for which `(epsilon_(0) h^(2)//pi me^(2) = 0.53 Å and m_(pi)//m_(e) = 264) pi-mesic` atoms might exist is

Options

A. `lt 105`
B. `gt 105`
C. `lt 37`
D. `gt 37`

Explanation

To solve the problem regarding the limit on the atomic number \( Z \) for which \( \pi \)-mesic atoms might exist, we will follow these steps:

### Step 1: Understand the Given Information
We have a \( \pi \)-meson of mass \( m_{\pi} \) and charge equal to that of an electron, which is in a circular orbit around a nucleus with atomic number \( Z \). The angular momentum is given by:
\[ L = \frac{h}{2\pi} \]
The radius of the nucleus is given by:
\[ R = 1.6 \times 10^{-15} Z^{1/3} \, \text{m} \]

### Step 2: Use the Angular Momentum Condition
From the Bohr model, the angular momentum \( L \) can also be expressed as:
\[ L = m_{\pi} v r \]
Setting this equal to the given angular momentum:
\[ m_{\pi} v r = \frac{h}{2\pi} \]

### Step 3: Use the Centripetal Force Condition
The centripetal force required to keep the \( \pi \)-meson in circular motion is provided by the Coulomb force:
\[ \frac{m_{\pi} v^2}{r} = \frac{Z e^2}{4 \pi \epsilon_0 r^2} \]
Rearranging gives:
\[ m_{\pi} v^2 = \frac{Z e^2}{4 \pi \epsilon_0 r} \]

### Step 4: Solve for the Radius \( r \)
From the angular momentum equation, we can express \( v \) in terms of \( r \):
\[ v = \frac{h}{2\pi m_{\pi} r} \]
Substituting this into the centripetal force equation:
\[ m_{\pi} \left(\frac{h}{2\pi m_{\pi} r}\right)^2 = \frac{Z e^2}{4 \pi \epsilon_0 r} \]
This simplifies to:
\[ \frac{h^2}{4 \pi^2 m_{\pi} r^2} = \frac{Z e^2}{4 \pi \epsilon_0 r} \]
Multiplying both sides by \( 4 \pi \epsilon_0 r^2 \):
\[ h^2 \epsilon_0 = Z e^2 \pi m_{\pi} r \]
Thus, we can express \( r \):
\[ r = \frac{h^2 \epsilon_0}{Z e^2 \pi m_{\pi}} \]

### Step 5: Substitute Known Values
We know:
\[ \frac{\epsilon_0 h^2}{\pi m_e e^2} = 0.53 \, \text{Å} = 0.53 \times 10^{-10} \, \text{m} \]
And the mass ratio:
\[ \frac{m_{\pi}}{m_e} = 264 \implies m_{\pi} = 264 m_e \]
Substituting into the radius equation:
\[ r = \frac{0.53 \times 10^{-10} \times m_e}{Z \cdot 264 \cdot e^2} \]

### Step 6: Compare with Nuclear Radius
We need \( r \) to be greater than the nuclear radius \( R \):
\[ \frac{0.53 \times 10^{-10}}{Z \cdot 264} > 1.6 \times 10^{-15} Z^{1/3} \]

### Step 7: Rearranging the Inequality
Rearranging gives:
\[ 0.53 \times 10^{-10} > 1.6 \times 10^{-15} Z^{1/3} \cdot Z \cdot 264 \]
This simplifies to:
\[ Z^{4/3} < \frac{0.53 \times 10^{-10}}{1.6 \times 10^{-15} \cdot 264} \]

### Step 8: Calculate the Limit for \( Z \)
Calculating the right side:
\[ Z^{4/3} < \frac{0.53 \times 10^{-10}}{4.224 \times 10^{-13}} \implies Z^{4/3} < 125.0 \]
Taking both sides to the power of \( \frac{3}{4} \):
\[ Z < 37.0 \]

### Conclusion
Thus, the limit on \( Z \) for which \( \pi \)-mesic atoms might exist is:
\[ \boxed{Z < 37} \]