An X-rays tube has a short wavelength end at `0.45A`. The voltage of tube is

Options

A. 450000 V
B. 9600 V
C. 27500 V
D. 60600 V

Answer

To find the voltage of an X-ray tube that emits radiation with a short wavelength of \(0.45 \, \text{Å}\) (angstrom), we can follow these steps:

### Step 1: Convert Wavelength to SI Units
The given wavelength is \(0.45 \, \text{Å}\). We need to convert this to meters:
\[ 1 \, \text{Å} = 10^{-10} \, \text{m} \]
Thus,
\[ \lambda = 0.45 \, \text{Å} = 0.45 \times 10^{-10} \, \text{m} = 4.5 \times 10^{-11} \, \text{m} \]

### Step 2: Calculate the Energy of the X-ray Photon
The energy \(E\) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) (Planck's constant) \(= 6.63 \times 10^{-34} \, \text{J s}\)
- \(c\) (speed of light) \(= 3 \times 10^{8} \, \text{m/s}\)

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^{8} \, \text{m/s})}{4.5 \times 10^{-11} \, \text{m}} \]

### Step 3: Perform the Calculation
Calculating the numerator:
\[ 6.63 \times 10^{-34} \times 3 \times 10^{8} = 1.989 \times 10^{-25} \, \text{J m} \]
Now, divide by the wavelength:
\[ E = \frac{1.989 \times 10^{-25}}{4.5 \times 10^{-11}} \approx 4.42 \times 10^{-15} \, \text{J} \]

### Step 4: Relate Energy to Voltage
The energy of the photon can also be expressed in terms of the voltage \(V\) of the tube:
\[ E = eV \]
where \(e\) (the charge of an electron) is approximately \(1.6 \times 10^{-19} \, \text{C}\).

### Step 5: Solve for Voltage
Rearranging the equation gives:
\[ V = \frac{E}{e} \]
Substituting the values:
\[ V = \frac{4.42 \times 10^{-15} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} \approx 27625 \, \text{V} \]
This can be rounded to:
\[ V \approx 27500 \, \text{V} \]

### Final Answer
The voltage of the X-ray tube is approximately \(27500 \, \text{V}\).

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