Question

If the radius of the first orbit of a hydrogen atom is $R$ the De Broglie wavelength of an electron in the third orbit is?

Answer 1

Hint: In the given question, we are required to find the De Broglie wavelength of an electron in the third orbit. So, we are given the radius of the first orbit of a hydrogen atom as $R$. We will use the equation stating the angular momentum of an electron as \[mvR = \dfrac{{nh}}{{2\pi }}\]. Then, we will substitute the value of n and find the required answer.

Complete step by step answer:
The angular momentum of an electron is,
\[mvR = \dfrac{{nh}}{{2\pi }}\] −−−−− (1)
Rewriting this equation we get,
\[v = \dfrac{{nh}}{{2\pi mR}}\]
De Broglie wavelength is,
$\lambda = \dfrac{h}{{mv}}$ −−−− (2)
Substituting the value of $v$ in (2)
\[ \Rightarrow \lambda = \dfrac{h}{{m \times \left( {\dfrac{{nh}}{{2\pi mR}}} \right)}}\]
$ \Rightarrow \lambda = \dfrac{{2\pi R}}{n}$
Wavelength of the electron in the third orbit is given by substituting $n = 3$.
$\therefore \lambda = \dfrac{{2\pi R}}{3}$

Therefore, the wavelength of the electron in the third orbit is $\dfrac{{2\pi R}}{3}$.

Additional information:De Broglie gave an explanation for the quantization of the electron's angular momentum in the hydrogen atom, which was hypothesized in Bohr's quantum theory, using the idea of the electron matter wave. When we suppose that an electron in a hydrogen atom behaves like a wave rather than a particle, the scientific explanation for the first Bohr quantization condition emerges clearly. For matter waves, the ratio between momentum and wavelength is $p = \dfrac{h}{\lambda }$ , while the relationship between energy and frequency is \[E{\text{ }} = {\text{ }}hf\] . The de Broglie wavelength is equal to $\dfrac{h}{p}$ , these are known as de Broglie relations.

Note: The first orbit of the hydrogen atom is also known as the Bohr orbit. It is clear from the equation above that as the radius of the orbit increases the wavelength also increases. As the wavelength increases the energy decreases because energy and wavelength are related by the equation, $E = \dfrac{{hc}}{\lambda }$.