De Broglie wavelength of an electron in hydrogen ground state

[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