Question

Q1 A certain ruby laser emits 1.00 –j- pulses of light whose wavelength is 694 nm. What is the minimum of cr+3 ions in the ruby?

Answer 1

Answer:

the minimum of cr+3 ions in the ruby is

$N = 3.5 \times 10^{18}$

Explanation:

As we know that energy due to one photon of light is given as

$E = \frac{hc}{\lambda}$

$E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{694 \times 10^{-9}}$

so we have

$E = 2.85 \times 10^{-19} J$

now for number of cr+3 ions we have

$N = \frac{E_{total}}{E}$

$N = \frac{1}{2.85 \times 10^{-19}}$

$N = 3.5 \times 10^{18}$

#Learn

Topic : Energy of photons

https://brainly.in/question/5384916