a. Calculate the de Broglie wavelength of the electron in the $\mathrm{n}=1,2,\mathrm{and}3$states of the hydrogen atom. Use the information in Table 38.2 .

b. Show numerically that the circumference of the orbit for each of these stationary states is exactly equal to n de Broglie wavelengths.

c. Sketch the de Broglie standing wave for the$\mathrm{n}=3$ orbit

The wavelength of the electron in the states of the hydrogen atom according to de Broglie is n =1,2 and 3

Let's begin with the de Broglie wavelength

$\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}$

${\mathrm{\lambda }}_1=\frac{\mathrm{h}}{{\mathrm{mv}}_1}$(a electron in the n=1 state)

localid="1651474181581" ${\mathrm{\lambda }}_1=\frac{6.63\times {10}^{-34}}{\left(9.11\times {10}^{-31}\right)\left(2.19\times {10}^6\right)}$(Table 38.2 can be used as a substitute)

localid="1650704330020" ${\mathrm{\lambda }}_1=0.332\mathrm{nm}$

localid="1650704348538" ${\mathrm{\lambda }}_2=\frac{\mathrm{h}}{{\mathrm{mv}}_2}$(a electron in the n=2 state)

localid="1651474203790" ${\mathrm{\lambda }}_2=\frac{6.63\times {10}^{-34}}{\left(9.11\cdot {10}^{-31}\right)\left(1.09\times {10}^6\right)}$(Table 38.2 can be used as a substitute)

localid="1650704334770" ${\mathrm{\lambda }}_2=0.667\mathrm{nm}$

${\mathrm{\lambda }}_3=\frac{\mathrm{h}}{{\mathrm{mv}}_3}$(a electron in the n=3 state)

localid="1651474233990" ${\lambda }_3=\frac{6.63\times {10}^{-34}}{\left(9.11\times {10}^{-31}\right)\left(0.73\times {10}^6\right)}$(Table 38.2 can be used as a substitute)

${\mathrm{\lambda }}_3=0.996\mathrm{nm}$

The orbit circumference for each stationary state is equal to n de Broglie wavelengths

The circumference of the orbit is

For state 1

For state 2

For state 3

The circumference of the orbit has been shown to be exactly equal to the n time of the de-Broglie wavelength.

De Broglie proposed that electrons operate as standing waves as they orbit the nucleus (stationary waves). Standing waves are formed when two waves travel in opposite directions and are one wavelength out of phase. As a result, the two waves superimpose and form a zero-amplitude standing wave.

The de-Broglie standing wave of the orbit is depicted below.

The standing wave diagram is shown above.