Question

a type of radiation has an energy of ( 4.70 \times 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. (square) nm (square \times 10^{square})

Explanation:

Step1: Recall the formula for photon energy

The energy of a photon is given by \( E = h
u \), and the relationship between frequency (\(
u\)) and wavelength (\(\lambda\)) is \( c=\lambda
u \), so we can combine these to get \( E=\frac{hc}{\lambda} \), where \( h = 6.626\times 10^{-34}\ \text{J·s} \) (Planck's constant), \( c = 3.00\times 10^{8}\ \text{m/s} \) (speed of light), and we need to solve for \(\lambda\): \( \lambda=\frac{hc}{E} \). First, convert the energy from kJ to J: \( E = 4.70\times 10^{-22}\ \text{kJ}\times1000\ \text{J/kJ}=4.70\times 10^{-19}\ \text{J} \).

Step2: Substitute values into the formula

Substitute \( h = 6.626\times 10^{-34}\ \text{J·s} \), \( c = 3.00\times 10^{8}\ \text{m/s} \), and \( E = 4.70\times 10^{-19}\ \text{J} \) into \( \lambda=\frac{hc}{E} \):
\[
\lambda=\frac{(6.626\times 10^{-34}\ \text{J·s})(3.00\times 10^{8}\ \text{m/s})}{4.70\times 10^{-19}\ \text{J}}
\]
First, calculate the numerator: \( (6.626\times 10^{-34})(3.00\times 10^{8})=1.9878\times 10^{-25}\ \text{J·m} \). Then divide by the energy: \( \lambda=\frac{1.9878\times 10^{-25}}{4.70\times 10^{-19}}\ \text{m} \approx 4.23\times 10^{-7}\ \text{m} \).

Step3: Convert meters to nanometers

Since \( 1\ \text{m}=10^{9}\ \text{nm} \), we convert \( \lambda \) to nanometers: \( \lambda = 4.23\times 10^{-7}\ \text{m}\times10^{9}\ \text{nm/m}=423\ \text{nm} \).

Answer:

\( 423 \)