An electron has a wavelength of 239 nm. Find its momentum. Planck’s constant is 6.63*10^-34?

Question

Jeremiah Adams · Accepted Answer

λ is equal to h/p. This equation contains three symbols: A) The particle's wavelength is represented by λ; B) Planck's constant is represented by h; and C) the particle's momentum is represented by p. λ = 239 x #10^-9# m
 h = 6.63 x #10^-34# kg #m^2# #s^-1#
 p = h /  λ p =   6.63 x #10^-34# kg #m^2# #s^-1# /  239 x #10^-9# m p = 0.0277 x #10^ -25# kg m #s^-1#

Serenity Allen · Answer

undefined To find the momentum of an electron, we can use the de Broglie wavelength formula:

$$ \text{wavelength} = \frac{h}{p} $$

Where:
- $ \text{wavelength} = 239 \, \text{nm} = 239 \times 10^{-9} \, \text{m} $
- $ h = 6.63 \times 10^{-34} \, \text{m}^2 \cdot \text{kg/s} $ (Planck's constant)
- $ p $ is the momentum of the electron.

Rearranging the formula to solve for momentum:

$$ p = \frac{h}{\text{wavelength}} $$

Substituting the given values:

$$ p = \frac{6.63 \times 10^{-34} \, \text{m}^2 \cdot \text{kg/s}}{239 \times 10^{-9} \, \text{m}} $$

$$ p = \frac{6.63 \times 10^{-34}}{239} \times 10^{-9} \, \text{kg} \cdot \text{m/s} $$

$$ p \approx 2.78 \times 10^{-27} \, \text{kg} \cdot \text{m/s} $$

So, the momentum of the electron is approximately $ 2.78 \times 10^{-27} \, \text{kg} \cdot \text{m/s} $.