Quantum Physics

Given Information:
Power of source, $P%20%3D%2060%5C%2CW$
Wavelength of monochromatic light, $%5Clambda%20%3D%20662.5%5C%2Cnm%20%3D%20662.5%20%5Ctimes%2010%5E%7B-9%7D%5C%2Cm$

Step-by-Step Explanation:

Step 1: Energy per photon clearly calculated:

The energy of each photon ( $E$ ) is given by:

$E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D$

where $h%20%3D%206.626%5Ctimes10%5E%7B-34%7D%5C%2CJs$ , and $c%20%3D%203%5Ctimes10%5E8%5C%2Cm%2Fs$ .

Substitute clearly:

$E%20%3D%20%5Cfrac%7B6.626%5Ctimes10%5E%7B-34%7D%5Ctimes%203%5Ctimes10%5E8%7D%7B662.5%5Ctimes10%5E%7B-9%7D%7D$

Simplifying clearly,

$E%20%5Capprox%20%5Cfrac%7B19.878%5Ctimes10%5E%7B-26%7D%7D%7B662.5%5Ctimes10%5E%7B-9%7D%7D%20%5Capprox%203%5Ctimes10%5E%7B-19%7D%5C%2CJ$

Step 2: Calculate clearly the number of photons emitted per second:

The number of photons emitted per second ( $n$ ) is given by power divided by energy per photon:

$n%20%3D%20%5Cfrac%7BP%7D%7BE%7D%20%3D%20%5Cfrac%7B60%7D%7B3%5Ctimes10%5E%7B-19%7D%7D%20%3D%202%5Ctimes10%5E%7B20%7D%5C%2Cs%5E%7B-1%7D$

Final Conclusion:
The number of photons emitted per second is clearly $2%5Ctimes10%5E%7B20%7D%5C%2Cs%5E%7B-1%7D$ .

High-Yield Trend

Chapter Questions
1 MCQs

Official Solution

Quantum Physics

High-Yield Trend

Chapter Questions 1 MCQs

Official Solution

Chapter Questions
1 MCQs