if the laser with a frequency of 4.558×10¹⁴ hz emits 6.49×10⁻¹ j of energy during a pulse, how many photons…

if the laser with a frequency of 4.558×10¹⁴ hz emits 6.49×10⁻¹ j of energy during a pulse, how many photons are emitted? report your answer in scientific notation to the correct number of sig figs by filling in the blanks below. do not include units in your answer.\n\nblank×10blank photons per pulse\n\nquestion 5\n\nin what region of the electromagnetic spectrum would you find a photon with an energy of 5.33×10⁻²³ j?\n\nfrequency (s⁻¹)\n10²⁰ 10¹⁸ 10¹⁶ 10¹⁴ 10¹² 10¹⁰ 10⁸ 10⁶ 10⁴\ngamma ray x - ray ultra - violet visible infrared microwave radio frequency\n10² 10⁰ 10² 10⁴ 10⁶ 10⁸ 10¹⁰ 10¹²\nwavelength (nm)\n\nmicrowave\nx - ray\ngamma\nvisible\nradio/tv\ninfrared\nultra - violet

if the laser with a frequency of 4.558×10¹⁴ hz emits 6.49×10⁻¹ j of energy during a pulse, how many photons are emitted? report your answer in scientific notation to the correct number of sig figs by filling in the blanks below. do not include units in your answer.\n\nblank×10blank photons per pulse\n\nquestion 5\n\nin what region of the electromagnetic spectrum would you find a photon with an energy of 5.33×10⁻²³ j?\n\nfrequency (s⁻¹)\n10²⁰ 10¹⁸ 10¹⁶ 10¹⁴ 10¹² 10¹⁰ 10⁸ 10⁶ 10⁴\ngamma ray x - ray ultra - violet visible infrared microwave radio frequency\n10² 10⁰ 10² 10⁴ 10⁶ 10⁸ 10¹⁰ 10¹²\nwavelength (nm)\n\nmicrowave\nx - ray\ngamma\nvisible\nradio/tv\ninfrared\nultra - violet

Answer

Explanation:

Step1: Recall the energy - photon formula

The energy of a single photon is given by $E = h\nu$, where $h = 6.626\times10^{-34}\text{ J}\cdot\text{s}$ (Planck's constant) and $\nu$ is the frequency. Given $\nu=4.558\times 10^{14}\text{ Hz}$, the energy of a single photon $E_{1}=h\nu=(6.626\times 10^{-34}\text{ J}\cdot\text{s})\times(4.558\times 10^{14}\text{ s}^{-1})$. [E_{1}=6.626\times4.558\times 10^{-34 + 14}\text{ J}=3.029\times 10^{-19}\text{ J}]

Step2: Calculate the number of photons

Let $n$ be the number of photons. We know that the total energy $E_{total}=6.49\times 10^{-3}\text{ J}$ and $E_{total}=nE_{1}$. So, $n=\frac{E_{total}}{E_{1}}=\frac{6.49\times 10^{-3}\text{ J}}{3.029\times 10^{-19}\text{ J}}$. [n=\frac{6.49}{3.029}\times10^{-3+ 19}\approx2.14\times 10^{16}]

Answer:

$2.14$; $16$

Explanation for second part:

We know that the energy of a photon is $E = h\nu$, so $\nu=\frac{E}{h}$. Given $E = 5.33\times 10^{-23}\text{ J}$ and $h = 6.626\times 10^{-34}\text{ J}\cdot\text{s}$, then $\nu=\frac{5.33\times 10^{-23}\text{ J}}{6.626\times 10^{-34}\text{ J}\cdot\text{s}}$. [ \nu=\frac{5.33}{6.626}\times10^{-23 + 34}\text{ s}^{-1}\approx8.04\times 10^{10}\text{ s}^{-1}] Looking at the electromagnetic - spectrum chart, a frequency of $8.04\times 10^{10}\text{ s}^{-1}$ falls in the microwave region.

Answer:

Microwave