Assume electrons are accelerated through a potential difference of 25 000 V inside a TV picture tube.What is the minimum wavelength that could be produced when the electrons strike the phosphor? (1 Å = 10 ^- ¹⁰ m) A)0.50 Å B)1.0 Å C)10 Å D)100 Å E)0.25 Å

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Quizplus · Accepted Answer

The minimum wavelength is given by the de Broglie wavelength equation:$$\lambda = \frac{h}{p}$$where h is Planck's constant and p is the momentum of the electron. The momentum of the electron is given by:$$p = \sqrt{2mK}$$where m is the mass of the electron and K is the kinetic energy of the electron. The kinetic energy of the electron is given by:$$K = eV$$where e is the charge of the electron and V is the potential difference through which the electron is accelerated. Substituting these equations into the de Broglie wavelength equation, we get:$$\lambda = \frac{h}{\sqrt{2meV}}$$Plugging in the values for h, m, e, and V, we get:$$\lambda = \frac{6.626 	imes 10^{-34} 	ext{ J s}}{\sqrt{2(9.109 	imes 10^{-31} 	ext{ kg})(1.602 	imes 10^{-19} 	ext{ C})(25000 	ext{ V})}}$$$$\lambda = 0.50 	imes 10^{-10} 	ext{ m} = 0.50 	ext{ Å}$$Therefore, the minimum wavelength that could be produced when the electrons strike the phosphor is 0.50 Å.

Assume Electrons Are Accelerated Through a Potential Difference of 25