Assume electrons are accelerated through a potential difference of 25000 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 ($\lambda_{min}$) that can be produced by electrons accelerated through a potential difference (V) can be calculated using the equation derived from the energy and momentum conservation principles, which is given by the formula $\lambda_{min} = \frac{h}{\sqrt{2meV}}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} m^2 kg / s$), $m$ is the mass of an electron ($9.109 \times 10^{-31} kg$), $e$ is the elementary charge ($1.602 \times 10^{-19} C$), and $V$ is the potential difference in volts. Substituting the given values and constants into this formula, we find that the minimum wavelength corresponds to approximately 0.50 Å, which is the result of the energy conversion from electrical potential energy to kinetic energy and then to photon energy as the electrons decelerate upon impact.

Assume Electrons Are Accelerated Through a Potential Difference of 25000