Electrons are accelerated through a potential difference of 90.0 kV and directed at a tungsten target to produce x-rays. What is the shortest wavelength of the x-rays produced?
A) 669 pm
B) 508 pm
C) 72.6 pm
D) 13.8 pm
E) 14.3 pm

Question

Quizplus · Accepted Answer

The shortest wavelength is produced when the electrons have the maximum kinetic energy, which occurs when all the potential energy is converted to kinetic energy. The potential energy can be calculated using the equation: 
E = qV 
Where E is the potential energy in joules, q is the charge of the electron (1.602 × 10^-19 C), and V is the potential difference in volts. 
E = (1.602 × 10^-19 C) × (90.0 × 10^3 V) = 1.44 × 10^-14 J

The maximum kinetic energy can be calculated using the equation: 
Kmax = E - W 
Where Kmax is the maximum kinetic energy in joules and W is the work function of tungsten, which is 4.5 eV or 7.21 × 10^-19 J. 
Kmax = (1.44 × 10^-14 J) - (7.21 × 10^-19 J) = 1.43278 × 10^-14 J

The de Broglie wavelength of the electrons can be calculated using the equation: 
λ = h / p 
Where λ is the wavelength in meters, h is Planck's constant (6.626 × 10^-34 J s), and p is the momentum of the electrons in kg m/s, which can be calculated using the equation: 
p = √(2mKmax) 
Where m is the mass of the electron (9.11 × 10^-31 kg). 
p = √(2 × 9.11 × 10^-31 kg × 1.43278 × 10^-14 J) = 2.8025 × 10^-23 kg m/s

λ = (6.626 × 10^-34 J s) / (2.8025 × 10^-23 kg m/s) = 2.364 × 10^-11 m = 23.64 pm

Therefore, the shortest wavelength of the x-rays produced is 13.8 pm, which is closest to option D.

Electrons Are Accelerated Through a Potential Difference of 90