Question

the hydrogen emission spectrum is shown below. what is the energy of the 410 nm emission line? (the speed of light in a vacuum is 3.00 × 10⁸ m/s, and planck’s constant is 6.626 × 10⁻³⁴ j•s.)
400
h
750 nm
410 434
486
656 nm

a. 4.85 × 10⁻¹⁹ j
b. 6.19 × 10²⁶ j
c. 2.06 × 10¹⁸ j
d. 7.32 × 10¹⁴ j

Explanation:

Step1: Recall the energy formula for a photon

The energy \( E \) of a photon is given by \( E = h
u \), where \( h \) is Planck's constant and \(
u \) is the frequency. Also, the relationship between frequency \(
u \), speed of light \( c \), and wavelength \( \lambda \) is \( c=\lambda
u \), so \(
u=\frac{c}{\lambda} \). Substituting \(
u \) into the energy formula gives \( E = \frac{hc}{\lambda} \).

Step2: Convert wavelength to meters

The wavelength \( \lambda = 410 \, \text{nm} \). Since \( 1 \, \text{nm} = 10^{-9} \, \text{m} \), we have \( \lambda = 410\times 10^{-9} \, \text{m}=4.10\times 10^{-7} \, \text{m} \).

Step3: Plug in the values into the energy formula

We know \( h = 6.626\times 10^{-34} \, \text{J}\cdot\text{s} \), \( c = 3.00\times 10^{8} \, \text{m/s} \), and \( \lambda = 4.10\times 10^{-7} \, \text{m} \).

\[

$$\begin{align*} E&=\frac{hc}{\lambda}\\ &=\frac{(6.626\times 10^{-34} \, \text{J}\cdot\text{s})(3.00\times 10^{8} \, \text{m/s})}{4.10\times 10^{-7} \, \text{m}}\\ &=\frac{1.9878\times 10^{-25} \, \text{J}\cdot\text{m}}{4.10\times 10^{-7} \, \text{m}}\\ &\approx 4.85\times 10^{-19} \, \text{J} \end{align*}$$

\]

Answer:

A. \( 4.85 \times 10^{-19} \, \text{J} \)