An electron (m_e = 9.11 *10^-31 kg) traveling with a velocity of 5.50 *10⁶ m/s has a De Broglie wavelength of approximately A) 0.0710 nm B) 0.130 nm C) 0.180 nm D) 0.230 nm E) 0.550 nm

Question

Quizplus · Accepted Answer

The De Broglie wavelength (λ) of an electron can be calculated using the formula:

λ = h/p

where h is the Planck's constant (6.626 x 10^-34 J*s) and p is the momentum of the electron, which can be calculated as:

p = m*v

where m is mass of the electron and v is its velocity.

Putting the values in the equation, we get:

p = (9.11 x 10^-31 kg) * (5.50 x 10^6 m/s) = 5.01 x 10^-24 kg m/s

λ = (6.626 x 10^-34 J*s) / (5.01 x 10^-24 kg m/s) = 1.321 x 10^-10 m

Converting this value to nanometers (nm), we get:

λ = 0.132 nm

Therefore, the closest option is B, which is 0.130 nm.

An Electron (Me = 9

An Electron (M_e = 9