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The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?

Amit Saxena , 10 Years ago
Grade upto college level
anser 1 Answers
Navjyot Kalra

Last Activity: 10 Years ago

Given λ’ = λ – 26 pm, V’ = 1.2 V Now, λ = hc/ev , λ’ = hc/ev’ Or λV = λ’V’ ⇒ λV = (λ – 26 * 10^-12)* 1.5 V ⇒ λ = 1.5 λ – 1.5 * 26 * 10^–12 ⇒ λ = 39* 10^-12/0.5 = 78 * 10^12 m V = hc/eλ = 6.63 * 3* 10^-34* 10^8/1.6 *10^-19 * 78* 10^-12 = 0.15937 * 10^5 = 15.93 * 10^* V = 1593 KV.

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