Question

1. a. using this equation: energy change = plank’s constant (h) x frequency (v), calculate the frequency of the first line in the hydrogen spectrum. plank’s constant is 6.626×10^(-34) joule·sec and the energy change is 3.03×10^(-19) joules. show your work! b. then, using the equation: speed of light (c) = wavelength (λ) x frequency (v), calculate the wavelength of a hydrogen energy level. (use the frequency that you calculated above and the speed of light is 2.9979×10^8 m/s). show your work! c. now, look back at the spectrum that you colored of the hydrogen bulb that you observed in class, what do you see near the wavelength that you calculated?

Explanation:

Step1: Rearrange energy - frequency formula

Given $E = h
u$, we can solve for $
u$ (frequency) as $
u=\frac{E}{h}$.

Step2: Substitute values for part a

Substitute $E = 3.03\times 10^{-19}\text{ J}$ and $h = 6.626\times 10^{-34}\text{ J}\cdot\text{s}$ into the formula:
$
u=\frac{3.03\times 10^{-19}\text{ J}}{6.626\times 10^{-34}\text{ J}\cdot\text{s}}\approx4.57\times 10^{14}\text{ Hz}$

Step3: Rearrange speed - wavelength - frequency formula

Given $c=\lambda
u$, we can solve for $\lambda$ (wavelength) as $\lambda=\frac{c}{
u}$.

Step4: Substitute values for part b

Substitute $c = 2.9979\times 10^{8}\text{ m/s}$ and $
u = 4.57\times 10^{14}\text{ Hz}$ into the formula:
$\lambda=\frac{2.9979\times 10^{8}\text{ m/s}}{4.57\times 10^{14}\text{ Hz}}\approx6.56\times 10^{-7}\text{ m} = 656\text{ nm}$

Step5: Answer for part c (conceptual)

Near a wavelength of $656\text{ nm}$ in the hydrogen spectrum, one would see a red - colored line (the Balmer - alpha line).

Answer:

a. $4.57\times 10^{14}\text{ Hz}$
b. $656\text{ nm}$
c. A red - colored line.