What is the de broglie wavelength of an electron (m = 9. 11 × 10-31 kg) moving at a velocity of 3. 0 × 107 m/s?
1 answer:
the de Broglie wavelength of the electron moving at a velocity of 3. 0 × 107 m/s is 2.4×10∧-11nm
lambda = h/mv
The de Broglie wavelength is defined as follows: lambda = h/mv, wherein the greek letter lambda represents the wavelength, h is Planck's consistent, m is the particle's mass and v is its pace. One may also explicit mv as the particle's momentum
according to wave-particle duality, the De Broglie wavelength is a wavelength manifested in all the gadgets in quantum mechanics which determines the chance density of locating the item at a given factor of the configuration area. The de Broglie wavelength of a particle is inversely proportional to its momentum
Learn more about de Broglie wavelength here: brainly.com/question/19052579
#SPJ4
You might be interested in
Answer:
0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂
Explanation:
In first place, the balanced reaction between Mg and O₂ is:
2 Mg + O₂ ⇒ 2 MgO
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:
- Mg: 2 moles
- O₂: 1 mole
- MgO: 2 moles
Then you can apply the following rule of three: if by reaction stoichiometry 2 moles of Mg produce 2 moles of MgO, 0.250 moles of Mg, how many moles of MgO will they form?
moles of MgO= 0.250
<u><em>0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂</em></u>