Question

1. what is the energy of light whose wavelength is 4.06 x 10^-11 m?

Explanation:

Step1: Recall the energy - wavelength formula

The energy of a photon is given by $E = h
u$, and since $
u=\frac{c}{\lambda}$, then $E=\frac{hc}{\lambda}$, where $h = 6.626\times10^{-34}\ J\cdot s$ (Planck's constant), $c = 3\times10^{8}\ m/s$ (speed of light in vacuum), and $\lambda$ is the wavelength of the photon.

Step2: Substitute the values for problem 11

For $\lambda = 6.4\times10^{-7}\ m$, we have $E=\frac{6.626\times10^{-34}\ J\cdot s\times3\times10^{8}\ m/s}{6.4\times10^{-7}\ m}$.
First, calculate the numerator: $6.626\times10^{-34}\times3\times10^{8}=19.878\times10^{-26}\ J\cdot m$.
Then, divide by the denominator: $E=\frac{19.878\times10^{-26}\ J\cdot m}{6.4\times10^{-7}\ m}=3.106\times10^{-19}\ J$.

Step3: Substitute the values for problem 12

For $\lambda = 4.06\times10^{-11}\ m$, $E=\frac{6.626\times10^{-34}\ J\cdot s\times3\times10^{8}\ m/s}{4.06\times10^{-11}\ m}$.
Calculate the numerator: $6.626\times10^{-34}\times3\times10^{8}=19.878\times10^{-26}\ J\cdot m$.
Divide by the denominator: $E=\frac{19.878\times10^{-26}\ J\cdot m}{4.06\times10^{-11}\ m}=4.9\times10^{-15}\ J$.

Answer:

1. $3.106\times10^{-19}\ J$
2. $4.9\times10^{-15}\ J$