Question

a type of radiation has an energy of 4.70 × 10⁻²² kj. note: reference the fundamental constants table for additional information. part: 0 / 2 part 1 of 2 what is the wavelength of the radiation? be sure your answer has the correct number of significant digits. 423 nm ×10

Explanation:

Step1: Recall the energy - wavelength formula

The formula that relates the energy (\(E\)) of a photon to its wavelength (\(\lambda\)) is \(E = \frac{hc}{\lambda}\), where \(h = 6.626\times10^{-34}\space J\cdot s\) (Planck's constant) and \(c=3.00\times 10^{8}\space m/s\) (speed of light in vacuum). We can re - arrange this formula to solve for \(\lambda\): \(\lambda=\frac{hc}{E}\)

First, we need to convert the energy from \(kJ\) to \(J\). Given \(E = 4.70\times 10^{-22}\space kJ\), since \(1\space kJ=1000\space J = 10^{3}\space J\), then \(E=4.70\times 10^{-22}\times10^{3}\space J=4.70\times 10^{-19}\space J\)

Step2: Substitute the values into the formula

Substitute \(h = 6.626\times 10^{-34}\space J\cdot s\), \(c = 3.00\times 10^{8}\space m/s\) and \(E=4.70\times 10^{-19}\space J\) into the formula \(\lambda=\frac{hc}{E}\)

\[

$$\begin{align*} \lambda&=\frac{6.626\times 10^{-34}\space J\cdot s\times3.00\times 10^{8}\space m/s}{4.70\times 10^{-19}\space J}\\ &=\frac{6.626\times3.00\times 10^{-34 + 8}}{4.70\times 10^{-19}}\space m\\ &=\frac{19.878\times 10^{-26}}{4.70\times 10^{-19}}\space m\\ &=\frac{19.878}{4.70}\times10^{-26 + 19}\space m\\ &\approx4.23\times 10^{-7}\space m \end{align*}$$

\]

Step3: Convert meters to nanometers

Since \(1\space m = 10^{9}\space nm\), then \(\lambda=4.23\times 10^{-7}\space m\times\frac{10^{9}\space nm}{1\space m}=423\space nm\)

Answer:

\(423\space nm\)