Question

How does the DeBroglie wavelength of the electron in the ground state of a hydrogen atom compare to the DeBroglie wavelength of the electron when it's in a 2p state? The DeBroglie wavelengths of the two electrons are equal. The DeBroglie wavelength of the electron in the ground state is greater than that of the electron in a 2p state. The DeBroglie wavelength of the electron in the ground state is less than that of the electron in a 2p state. There is no way to compare the DeBroglie wavelengths of these two electrons.

Answer 1

Answer:

Condition for orbit stability,

                                                $n\lambda = 2\pi r_{}n$

where

So,          $n\lambda = 2\pi n^{2}a_{0}$

or                                             $\lambda = 2\pi na_{0}$

where a₀ is the Bohr's radius.

Therefore, the debroblie's wavelength of the orbiting electron in a hydrogen atom is only depends on the principle quantum number n and does not dependent on the azomuthal quantum number l.

Here we have electron in the ground state, that means n = 1.

Thus, its debroblie's wavelength is             $\lambda = 2\pi (1)a_{0} = 2\pi a_{0}$

and for the electron in 2p state, n = 2

Therefore, its debroblie's wavelength is $\lambda = 2\pi (2)a_{0} = 4\pi a_{0}$

Hence, the electron in the 2p state has a greater debroblie's wavelength than the electron in the ground state.

Option (C) is correct.