Answer to Question #171469 in General Chemistry for Dawson

Question

An x-ray has a wavelength of 4.40 Å. Calculate the energy (in J) of one photon of this radiation. Enter your answer in scientific notation.

Answer 1

Let Energy in J be...... E(J)

$E(J)=h(j.s)×v(Hz)$

Where $v(Hz)=\frac{C(m/s)}{\lambda(m)}$

h = Planck's constant $6.63×10^{-34}J.s$

C= Speed of light $3.00×10^{8}m/s$

Convert the wavelength from Angstrom Å to meter (m) $4.40 × \frac{1}{10^{10}}$ = $4.40×10^{-10}$

Therefore

$E(J) = 6.63×10^{-34}(\frac{3.00×10^8}{4.40×10^{-10}})$

$E(J) = 4.52×10^{-16}$