Question

in an experiment, a photon with a frequency of 7 × 10¹⁴ s⁻¹ was absorbed by a sample. what is the approximate energy of this photon? a 5 × 10⁻¹⁹ j b 1 × 10⁻¹⁵ j c 2 × 10⁶ j d 1 × 10⁴⁸ j

Explanation:

Step1: Recall the formula for photon energy

The energy of a photon is given by \(E = h
u\), where \(h\) is Planck's constant (\(h=6.626\times 10^{-34}\space J\cdot s\)) and \(
u\) is the frequency of the photon.

Step2: Substitute the given values into the formula

Given \(
u = 7\times 10^{14}\space s^{-1}\) and \(h = 6.626\times 10^{-34}\space J\cdot s\). Then \(E=(6.626\times 10^{-34}\space J\cdot s)\times(7\times 10^{14}\space s^{-1})\).
Using the rule of exponents \(a^m\times a^n=a^{m + n}\), we have \(E=(6.626\times7)\times10^{-34 + 14}\space J\).
\(6.626\times7 = 46.382\approx46\), and \(-34 + 14=-20\). So \(E = 46\times10^{-20}\space J=4.6\times 10^{-19}\space J\approx5\times 10^{-19}\space J\).

Answer:

A. \(5\times 10^{-19}\space J\)