Question

The de Broglie wavelength of the electron in the ground state of the hydrogen atom is ……………… (radius of the first orbit of hydrogen atom $= 0.53\mathop {\text{A}}\limits^0$)
(A) $3.33\mathop {\text{A}}\limits^0$
(B) $1.67\mathop {\text{A}}\limits^0$
(C) $0.53\mathop {\text{A}}\limits^0$
(D) $1.06\mathop {\text{A}}\limits^0$

Answer 1

In Bohr’s model, the angular momentum of an electron around a nucleus must be an integer multiple of the reduced Planck’s constant (which is Planck's constant over 2 pi). The ground state signifies that that integer is equal to 1.

Formula used: In this solution we will be using the following formulae;
$p = \dfrac{h}{{2\pi r}}$ where $p$ is the momentum of the electron of an atom in an orbit, $h$ is the Planck’s constant, and $r$ is the radius of the orbit.
$\lambda = \dfrac{h}{p}$ where $\lambda$ is the de Broglie wavelength of a wave, $p$ and $h$ remains the momentum and Planck’s constant respectively.

Complete Step-by-Step Solution:
We are to find the de Broglie wavelength of an electron which is at the ground state of a hydrogen atom.
In the Bohr model of the atom, the electrons are considered to be revolving around a nucleus at different by unchanging orbit. Hence, these electrons have a velocity and a momentum.
The momentum is given by the equation
$p = \dfrac{h}{{2\pi r}}$ where $h$ is the Planck’s constant, and $r$ is the radius of the orbit in which the electron revolves around the nucleus.
Now, according to de Broglie, any matter with a momentum has a wavelength which is inversely proportional to that momentum. i.e.,
$\lambda = \dfrac{h}{p}$ where $\lambda$ is the de Broglie wavelength.
Rearranging the above equation to make $p$ subject of the formula, we get
$p = \dfrac{h}{\lambda }$ hence substituting this expression for $p$ in $p = \dfrac{h}{{2\pi r}}$, we have
$\dfrac{h}{\lambda } = \dfrac{h}{{2\pi r}}$
Hence, by cancellation of $h$ and inversion, we get
$\lambda = 2\pi r$
Hence, by substitution of given value,
$\lambda = 2\pi \left( {0.53\mathop {\text{A}}\limits^0 } \right) = 3.33\mathop {\text{A}}\limits^0$

Hence, the correct option is A

Note: For clarity, the equation $p = \dfrac{h}{{2\pi r}}$ can be derived from the Bohr statement in his model of the atom that the angular momentum of an atom must be equal to an integer multiple of the reduced Planck’s constant i.e.
$L = mvr = \dfrac{{nh}}{{2\pi }}$
For ground state $n = 1$, hence,
$p = \dfrac{h}{{2\pi r}}$ (since $p = mv$)