If there times of `lambda_(min)` of continuous `X-`ray spectrum of target metal at `40 kV` is some as the wavelength of `K_(alpha)` line of this metal at `30 kV`, then determine the atomic number of the target metal.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between the minimum wavelength and the energy of the X-rays
The minimum wavelength (\( \lambda_{min} \)) of the continuous X-ray spectrum is given by the formula:
\[ \lambda_{min} = \frac{hc}{E} \]
where \( E \) is the energy of the electrons, which can be expressed in terms of the voltage (V) applied:
\[ E = eV \]
Here, \( e \) is the charge of an electron, and \( V \) is the voltage in volts.

### Step 2: Calculate \( \lambda_{min} \) at 40 kV
Given that the voltage is 40 kV (or 40,000 volts), we can substitute this into the equation:
\[ \lambda_{min} = \frac{hc}{e \cdot 40,000} \]
Using the values \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \), and converting to appropriate units, we find:
\[ \lambda_{min} = \frac{1240 \, \text{eV} \cdot \text{nm}}{40} = 0.031 \, \text{nm} = 0.31 \, \text{Å} \]

### Step 3: Relate \( \lambda_{min} \) to the Kα line
According to the problem, three times the minimum wavelength at 40 kV is equal to the wavelength of the Kα line at 30 kV:
\[ 3 \lambda_{min} = \lambda_{K\alpha} \]
Substituting the value of \( \lambda_{min} \):
\[ 3 \times 0.31 \, \text{Å} = \lambda_{K\alpha} \]
Thus,
\[ \lambda_{K\alpha} = 0.93 \, \text{Å} \]

### Step 4: Calculate the energy of the Kα line
The energy of the Kα line can be expressed as:
\[ E_{K\alpha} = \frac{hc}{\lambda_{K\alpha}} \]
Substituting \( \lambda_{K\alpha} \):
\[ E_{K\alpha} = \frac{1240 \, \text{eV} \cdot \text{nm}}{0.093 \, \text{nm}} = \frac{1240}{0.093} \approx 13333.33 \, \text{eV} \]

### Step 5: Relate the energy of Kα to the atomic number
The energy of the Kα line is also given by:
\[ E_{K\alpha} = \frac{3}{4} \cdot 13.6 (Z - 1)^2 \]
Setting the two expressions for \( E_{K\alpha} \) equal:
\[ 13333.33 = \frac{3}{4} \cdot 13.6 (Z - 1)^2 \]

### Step 6: Solve for \( Z \)
Rearranging gives:
\[ (Z - 1)^2 = \frac{13333.33 \cdot 4}{3 \cdot 13.6} \]
Calculating the right side:
\[ (Z - 1)^2 \approx 1308 \]
Taking the square root:
\[ Z - 1 \approx \sqrt{1308} \approx 36 \]
Thus:
\[ Z \approx 37 \]

### Final Result
The atomic number of the target metal is approximately **37**.

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