To find the minimum accelerating voltage required for hard X-rays with a minimum wavelength of \(10^{-11} \, m\), we can use the relationship between energy and wavelength for photons, as well as the concept of the stopping potential in an X-ray machine.
### Step-by-Step Solution:
1. **Understand the relationship between energy and wavelength**:
The energy \(E\) of a photon can be expressed in terms of its wavelength \(\lambda\) using the equation:
\[
E = \frac{hc}{\lambda}
\]
where:
- \(h\) is Planck's constant (\(6.625 \times 10^{-34} \, J \cdot s\)),
- \(c\) is the speed of light (\(3 \times 10^{8} \, m/s\)),
- \(\lambda\) is the wavelength.
2. **Substitute the given wavelength**:
We are given that the minimum wavelength \(\lambda\) is \(10^{-11} \, m\). Substituting this into the energy equation:
\[
E = \frac{(6.625 \times 10^{-34} \, J \cdot s)(3 \times 10^{8} \, m/s)}{10^{-11} \, m}
\]
3. **Calculate the energy**:
Performing the calculation:
\[
E = \frac{(6.625 \times 3) \times 10^{-34 + 8 + 11}}{1} = \frac{19.875 \times 10^{-15}}{1} \, J
\]
\[
E \approx 1.9875 \times 10^{-14} \, J
\]
4. **Relate energy to voltage**:
The energy of the electrons accelerated through a potential \(V\) is given by:
\[
E = eV
\]
where \(e\) is the charge of an electron (\(1.6 \times 10^{-19} \, C\)). Thus, we can express the voltage as:
\[
V = \frac{E}{e}
\]
5. **Substitute the energy value**:
Now substituting the calculated energy into the voltage equation:
\[
V = \frac{1.9875 \times 10^{-14} \, J}{1.6 \times 10^{-19} \, C}
\]
6. **Calculate the voltage**:
Performing the division:
\[
V \approx 124.22 \times 10^{5} \, V = 124.22 \, kV
\]
7. **Conclusion**:
The accelerating voltage for the X-ray machine should be less than this calculated value. Therefore, the minimum accelerating voltage required for the X-ray machine is approximately \(124 \, kV\).
### Final Answer:
The accelerating voltage for electrons in the X-ray machine should be less than \(124 \, kV\).
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