Question

the absorption spectrum of neon has a line at 633 nm. what is the energy of this 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.)

a. 3.14 × 10⁻²⁸ j
b. 3.18 × 10²⁷ j
c. 3.14 × 10⁻¹⁹ j
d. 3.18 × 10¹⁸ j

Explanation:

Step1: Recall the formula for energy of 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, we get \( E = \frac{hc}{\lambda} \).

Step2: Convert wavelength to meters

The wavelength \( \lambda = 633 \, \text{nm} \). Since \( 1 \, \text{nm} = 10^{-9} \, \text{m} \), we have \( \lambda = 633\times 10^{-9} \, \text{m}=6.33\times 10^{-7} \, \text{m} \).

Step3: Substitute values into the formula

We know \( h = 6.626\times 10^{-34} \, \text{J·s} \), \( c = 3.00\times 10^{8} \, \text{m/s} \), and \( \lambda = 6.33\times 10^{-7} \, \text{m} \). Plugging these into \( E=\frac{hc}{\lambda} \):

\[

$$\begin{align*} E&=\frac{(6.626\times 10^{-34} \, \text{J·s})(3.00\times 10^{8} \, \text{m/s})}{6.33\times 10^{-7} \, \text{m}}\\ &=\frac{1.9878\times 10^{-25} \, \text{J·m}}{6.33\times 10^{-7} \, \text{m}}\\ &\approx 3.14\times 10^{-19} \, \text{J} \end{align*}$$

\]

Answer:

C. \( 3.14 \times 10^{-19} \, \text{J} \)