Question 42.5: De Broglie’s Model of Hydrogen If a hydrogen atom’s energy i......
De Broglie’s Model of Hydrogen
If a hydrogen atom’s energy is – 1.51 eV, what is the de Broglie wavelength of the atomic electron?
INTERPRET and ANTICIPATE
Use the energy-level diagram to find the principal quantum number. The quantum number determines the radius of the circular wire. (In Bohr’s model, this is the radius of the electron’s orbit.) The wavelength is determined by the quantum number and the radius of the circular wire.
SOLVE
The energy-level diagram (Fig. 42.7) shows which quantum number corresponds to an energy of -1.51 eV.
n = 3
Find the orbital radius from Equation 42.7, where r_B is the Bohr radius and n = 3.
\begin{aligned}& r=n^2 r_{ B } \quad \quad (42.7)\\& r=\left(3^2\right)\left(5.29 \times 10^{-11} m \right)=4.76 \times 10^{-10} m \end{aligned}The de Broglie wavelength is given by Equation 42.13, \lambda_n = 2\pi r/n.
\begin{aligned}& \lambda_3=\frac{2 \pi\left(4.76 \times 10^{-10} m \right)}{3} \\ & \lambda_3=9.97 \times 10^{-10} m \\& \lambda_3=0.997 nm\end{aligned}CHECK AND THINK
This problem required a reinterpretation of Bohr’s model. Instead of thinking of r as the radius of the electron’s orbit, we think of it as the radius of the wire to which the “waving” electron is confined.
