Question

determine the energy for the visible light in joules.
a 1.989 x 10³³
b 1.989 x 10⁻³³
c 1.989 x 10⁻²¹
d 1.989 x 10²¹

Explanation:

Step1: Recall the formula for energy of a photon

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 light. Also, \(
u=\frac{c}{\lambda}\), where \(c = 3\times10^{8}\space m/s\) (speed of light) and \(\lambda\) is the wavelength. For visible light, assume a wavelength \(\lambda = 1\times10^{-6}\space m\) (approximate value for calculation purposes as the problem does not specify a particular visible - light wavelength; the formula structure is key). Then \(
u=\frac{3\times 10^{8}}{1\times10^{-6}}=3\times10^{14}\space Hz\).

Step2: Calculate the energy

Using \(E = h
u\), substitute \(h = 6.626\times10^{-34}\space J\cdot s\) and \(
u=3\times 10^{14}\space Hz\). \(E=(6.626\times 10^{-34})\times(3\times10^{14})\). Using the rule of exponents \(a^{m}\times a^{n}=a^{m + n}\), we have \(E=(6.626\times3)\times10^{-34 + 14}\). \(6.626\times3 = 19.878\approx19.89\), and \(19.89\times10^{-20}=1.989\times10^{-21}\space J\) (scientific - notation adjustment: \(a\times10^{n}\) where \(1\leqslant a<10\)).

Answer:

C. \(1.989\times10^{-21}\)