Finding Broglie wavelength after acceleration

[Physicsq]

Homework Statement

A free electron initially moves along the positive x-axis direction with the de Broglie wavelength of 0.20 nm. If this election is further accelerated by 100 eV along the positive x-axis direction, what is the de Broglie wavelength of the electron after the acceleration?

Relevant Equations

λ=h/p
KE = 1/2mv^2

KE = 100eV x (1.6x10^-19)

v(new) = √(2KE/9.1×10^-31 )
= 5.9x10^6m/sv(original) = h/λm
= 6.6x10^-34 / (0.2x10^-9 x 9.1x10^-31)
= 3.64x10^6m/s

 

[ZapperZ]

1

That’s it?

Do you not know how to find the new KE after it has been accelerated?

Zz.

 

[Physicsq]

New KE will be 100eV x (1.6x10^-19)?

Do I add the velocity from the new KE to the current velocity traveling at 0.2nm?

 

[ZapperZ]

New KE will be 100eV x (1.6x10^-19)?

Do I add the velocity from the new KE to the current velocity traveling at 0.2nm?

From lessons on kinematics, you had this:

v(t) = v₀ + at

where v(t) is the velocity at time t, a is the acceleration, and v₀ is the initial velocity. What that equation say is that if you accelerate a particle over time t, then its final velocity will be the initial velocity added to whatever velocity it has gained from acceleration.

Does that answer your question?

Zz.