Electrons are accelerated through a potential difference of 90.0 kV and directed at a tungsten target to produce x-rays.What is the cutoff frequency in the x-ray spectrum produced? A) 2.17 $\times$ 10¹⁹ Hz B) 2.10 $\times$ 10¹⁹ Hz C) 1.17 $\times$ 10²¹ Hz D) 9.25 $\times$ 10²¹ Hz E) 6.20 $\times$ 10¹⁵ Hz

Question

Quizplus · Accepted Answer

The cutoff frequency is given by the equation f = (1/2π) * √(2e/m) * √(V), where e is the charge of an electron, m is the mass of an electron, and V is the potential difference. Plugging in the given values, we get f = (1/2π) * √(2 * 1.602 * 10^-19 C / 9.109 * 10^-31 kg) * √(90.0 * 10^3 V) = 2.17 * 10^19 Hz.

Electrons Are Accelerated Through a Potential Difference of 90 ×\times× 1019 Hz B) 2

Electrons Are Accelerated Through a Potential Difference of 90 $\times$ 10¹⁹ Hz
B) 2