MCQ
If the cathode-anode potential difference in an X-ray tube be $10 \mathrm{~V}$, then the maximum energy of X-ray photon can be
- A$10 \mathrm{~J}$
- B$10 \mathrm{MeV}$
- ✓$10 \mathrm{MeV}$
- D$10 \mathrm{KeV}$
✓
Answer
Correct option: C.
$10 \mathrm{MeV}$
Since $\lambda_{\min }=\frac{12375}{V} \mathring A=\frac{12375}{10^5} \mathring A=0.123 \mathring A$
$E_{\max }=\frac{h c}{\lambda_{\min }}$On putting the values. $E_{\max } \cong 10^{-1} \mathrm{MeV}$.
$E_{\max }=\frac{h c}{\lambda_{\min }}$On putting the values. $E_{\max } \cong 10^{-1} \mathrm{MeV}$.
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