The potential difference between the target and the cathode of an X-ray tube is 50 kV. The current in the tube is 20 mA. Only 1\% of the total energy is emitted as X-rays. (Take hc = 1240 eV nm )

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λ min ​ = 50 × 1 0 3 1240 ​ = 24.8 × 1 0 − 3 nm f max ​ = λ min ​ c ​ = 1240 × 1 0 − 9 3 × 1 0 8 × 50 × 1 0 3 ​ ≈ 1.2 × 1 0 19 Hz , so (A). Heat = V I − P xray ​ = 50 × 1 0 3 × 20 × 1 0 − 3 − 10 = 990 W , not 900 W , so (B) is false. 0.2 nm > λ min ​ = 0.0248 nm , so it can be emitted, (C). d t d N ​ = e I ​ = 1.6 × 1 0 − 19 20 × 1 0 − 3 ​ = 1.25 × 1 0 17 / sec , so ( D ) .

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