Question

1. what is the wavelength of a photon that has an energy of 2.52 x 10^(-19) j? identify the type of radiation. 7. a photon of light has an energy of 3.82 x 10^(-13) j. calculate the frequency & identify the type of radiation.

Explanation:

Step1: Recall energy - wavelength formula

The energy of a photon is given by $E = h\frac{c}{\lambda}$, where $E$ is energy, $h = 6.626\times10^{-34}\ J\cdot s$ (Planck's constant), $c= 3.0\times 10^{8}\ m/s$ (speed of light) and $\lambda$ is wavelength. We can re - arrange it to solve for $\lambda$: $\lambda=\frac{hc}{E}$.

Step2: Substitute values for first question

For $E = 2.52\times 10^{-19}\ J$, $h = 6.626\times10^{-34}\ J\cdot s$ and $c = 3.0\times 10^{8}\ m/s$.
$\lambda=\frac{(6.626\times 10^{-34}\ J\cdot s)\times(3.0\times 10^{8}\ m/s)}{2.52\times 10^{-19}\ J}$
$\lambda=\frac{19.878\times 10^{-26}\ J\cdot m}{2.52\times 10^{-19}\ J}=7.89\times 10^{-7}\ m = 789\ nm$. This is in the visible light region (400 - 700 nm is visible, but near - infrared starts around 700 nm, so it's near - infrared).

Step3: Recall energy - frequency formula for second question

The energy of a photon is also given by $E = h
u$, where $
u$ is frequency. We can solve for $
u$: $
u=\frac{E}{h}$.

Step4: Substitute values for second question

For $E = 3.82\times 10^{-13}\ J$ and $h = 6.626\times10^{-34}\ J\cdot s$.
$
u=\frac{3.82\times 10^{-13}\ J}{6.626\times10^{-34}\ J\cdot s}=5.76\times 10^{20}\ Hz$. This is in the gamma - ray region (frequencies above $10^{19}\ Hz$ are gamma - rays).

Answer:

1. Wavelength for first photon: $789\ nm$, type of radiation: near - infrared
2. Frequency for second photon: $5.76\times 10^{20}\ Hz$, type of radiation: gamma - ray