A particle (m = 2.7 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s2 for 5.0 s. What is its de Broglie wavelength λ at the end of this period?
Please help, I understand that v=at so you need to do 2.4 multiplied 5 and I got 12. Meaning v=12x10^7 Then I did h/mv which is (6.626x10-34)/(2.7x10^-28*12x10^7) and I finally got 2.045x10^-14 but this is still wrong and I don't understand why.
A particle (m = 2.7 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4...
A particle (m = 2.7 * 10-28 kg) starting from rest, experiences an acceleration of 2.4 x 107 m/s2 for 5.0 s. What is its de Broglie wavelength 1 at the end of this period? XE = Number 2.045e-14 Units the tolerance is +/-2%
A particle (m = 3.0 × 10-28 kg) starting from rest, experiences an acceleration of 2.4 × 107 m/s2 for 5.0 s. What is its de Broglie wavelength λ at the end of this period?
A particle (m = 3.0 × 10-28 kg) starting from rest, experiences an acceleration of 2.4 × 107 m/s2 for 5.0 s. What is its de Broglie wavelength λ at the end of this period?
1.In the Compton effect, an X-ray photon of wavelength 0.23 nm is incident on a stationary electron. Upon collision with the electron, the scattered X-ray photon continues to travel in the same direction as the incident photon. 1.What is the wavelength λ' of the scattered photon? λ' = 2.A particle (m = 3.7 × 10-28 kg) starting from rest, experiences an acceleration of 2.4 × 107 m/s2 for 5.0 s. What is its de Broglie wavelength λ at the end...
2). Rough estimation of the diameter of an aluminum atom: Given aluminum density as p-2.7*10 kg/m', atomic weight is 27 (i.e. the mass of 1 mole of aluminum atoms is 27g. Imole 6.02x10 ). Assume atoms of aluminum metal are spherical with spaces in between the aluminum atoms take up 74.1% of the volume of the material. What is the diameter of an aluminum atom? (Note: Volume of a sphere with radius ris: V-(4/3)xr) 4. (20") (1). In E-beam lithography,...