Multiple Choice
What is the de Broglie wavelength of an electron moving at a speed of 1.0 × 10^8 m/s?
9. Quantum Mechanics
De Broglie Wavelength
Struggling with General Chemistry?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
What is the de Broglie wavelength of an electron traveling at 1.12 x 10^5 m/s, given that the mass of an electron is 9.11 x 10^-31 kg?
A
7.52 x 10^-10 m
B
6.52 x 10^-9 m
C
5.52 x 10^-10 m
D
6.52 x 10^-10 m
Verified step by step guidance1
Start by recalling the de Broglie wavelength formula: \( \lambda = \frac{h}{mv} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ Js}) \), \( m \) is the mass of the particle, and \( v \) is the velocity of the particle.
Identify the given values: the mass of the electron \( m = 9.11 \times 10^{-31} \text{ kg} \) and the velocity \( v = 1.12 \times 10^{5} \text{ m/s} \).
Substitute the known values into the de Broglie wavelength formula: \( \lambda = \frac{6.626 \times 10^{-34} \text{ Js}}{(9.11 \times 10^{-31} \text{ kg})(1.12 \times 10^{5} \text{ m/s})} \).
Perform the multiplication in the denominator: \( (9.11 \times 10^{-31} \text{ kg}) \times (1.12 \times 10^{5} \text{ m/s}) \).
Divide Planck's constant by the result from the previous step to find the de Broglie wavelength \( \lambda \).
0:58mWatch next
Master De Broglie Wavelength Formula with a bite sized video explanation from Jules
Start learningRelated Videos
Related Practice
De Broglie Wavelength practice set
Problem sets built by lead tutorsExpert video explanations