De Broglie Wavelength of an electron

[Janet]

1. The velocity of the electron in the ground state of the hydrogen atom is 2.6 x 10^8 m/s. What is the wavelength of this electron in meters?2. De Broglie's equation: lamda = h/p
p=mv

The Attempt at a Solution

...

(6.626 x 10^-34) / (2.6 x 10^8 x 9.11 x 10^-31)

= 2.798 x 10^-12 meters

This isn't the right answer...obviously...[/B]

 

[e.bar.goum]

Quickly (sorry, posting and running), and with wolfram alpha, I get 1.7 * 10^-12 m. Which doesn't sound too far off to me (and neither does your answer), considering 2.6 * 10^8 m/s is about 500 keV of kinetic energy for an electron.

Why do you suspect your answer wrong?

 

[Simon Bridge]

Is the electron relativistic enough to make a difference?
Is the wavelength consistent with DeBroglie's hypothesis about the allowed energies of the electron?

 

[ehild]

1. The velocity of the electron in the ground state of the hydrogen atom is 2.6 x 10^8 m/s. What is the wavelength of this electron in meters?

Check the problem text. The speed of the electron is much less than the speed of light in the ground state of the H atom.

ehild

 

[HalfLostAndSearching]

There seems to be a mistake in the calculation. The correct equation to use for the de Broglie wavelength of an electron is λ = h/mv, where h is Planck's constant, m is the mass of the electron, and v is its velocity. Therefore, the correct calculation would be:

λ = (6.626 x 10^-34 m^2 kg/s) / (9.11 x 10^-31 kg) * (2.6 x 10^8 m/s)

= 2.42 x 10^-10 meters

This result is in agreement with the expected de Broglie wavelength for an electron in the ground state of a hydrogen atom. It is important to note that the de Broglie wavelength is a fundamental property of particles, and it can provide insight into their wave-like nature. In this case, the short wavelength of the electron suggests that it has a high probability of being found near the nucleus of the hydrogen atom.