Step 1: Calculate electron energy.
Voltage = 30 kV = \(30{,}000\) eV.
10% of this energy produces the photon:
\[
E = 0.10 \times 30{,}000 = 3000\ \text{eV}
\]
Step 2: Use the photon energy formula.
\[
E = \frac{hc}{\lambda}
\]
\[
\lambda = \frac{hc}{E}
\]
Step 3: Substitute values.
\[
\lambda = \frac{(4.14 \times 10^{-15})(3 \times 10^{8})}{3000}
\]
Convert to Å (1 Å = \(10^{-10}\) m):
\[
\lambda = \frac{1.242 \times 10^{-6}}{3000} = 4.14 \times 10^{-10}\ \text{m}
\]
\[
\lambda = 4.14\ \text{\AA}
\]
Step 4: Conclusion.
Wavelength of the photon = 4.14 Å.