Find the (a) maximum frequency and (b) minimum wave-length of X-rays produced by 30 kV electrons. Given, `h=6.63xx10^(-34)Js`.

Explanation

To solve the problem of finding the maximum frequency and minimum wavelength of X-rays produced by 30 kV electrons, we will follow these steps:

### Step 1: Calculate the Energy of the Electrons
The energy \( E \) of the electrons can be calculated using the formula:
\[ E = qV \]
where:
- \( q \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \))
- \( V \) is the potential difference (30 kV or \( 3 \times 10^4 \, \text{V} \))

Substituting the values:
\[ E = (1.6 \times 10^{-19} \, \text{C}) \times (3 \times 10^4 \, \text{V}) = 4.8 \times 10^{-15} \, \text{J} \]

### Step 2: Calculate the Maximum Frequency
The maximum frequency \( \nu \) of the X-rays can be found using the relation:
\[ E = h\nu \]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)). Rearranging the equation gives:
\[ \nu = \frac{E}{h} \]

Substituting the values:
\[ \nu = \frac{4.8 \times 10^{-15} \, \text{J}}{6.63 \times 10^{-34} \, \text{Js}} \approx 7.24 \times 10^{18} \, \text{Hz} \]

### Step 3: Calculate the Minimum Wavelength
The minimum wavelength \( \lambda \) can be calculated using the relationship between wavelength and frequency:
\[ \lambda = \frac{c}{\nu} \]
where \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

Substituting the values:
\[ \lambda = \frac{3 \times 10^8 \, \text{m/s}}{7.24 \times 10^{18} \, \text{Hz}} \approx 4.14 \times 10^{-11} \, \text{m} \]

To convert this to Angstroms (1 Angstrom = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 0.414 \, \text{Å} \]

### Final Answers
(a) Maximum frequency \( \nu \approx 7.24 \times 10^{18} \, \text{Hz} \)

(b) Minimum wavelength \( \lambda \approx 0.414 \, \text{Å} \)

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